 QUESTION

# Solve the quadratic equation for y: $\text{5}{{\text{y}}^{\text{2}}}\text{+5y-10=0}$

Hint: We will be using the concept of quadratic equation to tackle this question. Formula used for solving quadratic equation of the form $\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}$ is $\text{( }\!\!\alpha\!\! \text{ , } \!\!\beta\!\!\text{ )}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Before proceeding with the question, we should know about quadratic equations. Quadratic equations are the polynomial equations of degree 2 in one variable of type $\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}$ where a, b, c, $\in R$ and $\text{a}\ne \text{0}$. It is the general form of a quadratic equation where a is called the leading coefficient and c is called the absolute term. The values of x satisfying the quadratic equation are the roots of the quadratic equation $\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}$.
The quadratic equation will always have two roots. The nature of roots may be either real or imaginary.
Formula for solving quadratic equation of the form $\text{f(x)=a}{{\text{x}}^{\text{2}}}\text{+bx+c}$ is $\text{( }\!\!\alpha\!\!\text{ , }\!\!\beta\!\!\text{ )}=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}.....(1)$
So here in the question the variable is y so the general equation will become
$\text{a}{{\text{y}}^{\text{2}}}\text{+by+c=0}......\text{(2)}$
And the given equation is $\text{5}{{\text{y}}^{\text{2}}}\text{+5y-10=0}.....\text{(3)}$.
Now comparing equation (2) and equation (3) we get, $\text{a=5}$, $\text{b=5}$ and $\text{c =}\,-10$.
Now substituting the values of a, b and c in equation (1) we get,
$\text{y}=\dfrac{-5\pm \sqrt{{{5}^{2}}-4\times 5\times (-10)}}{2\times 5}.....(4)$
Now simplifying and rearranging equation (4) we get,
$\text{y}=\dfrac{-5\pm \sqrt{25+200}}{10}.....(5)$
Solving for y in equation (5) we get,
\begin{align} & \text{y}=\dfrac{-5\pm \sqrt{225}}{10} \\ & \text{y}=\dfrac{-5\pm 15}{10} \\ \end{align}
We get two values of y which are ${{\text{y}}_{1}}=\dfrac{-5-15}{10}=\dfrac{-20}{10}=-2$ and ${{\text{y}}_{2}}=\dfrac{-5+15}{10}=\dfrac{10}{10}=1$.
Hence we get -2 and 1 as the answer.

Note: An alternate method for solving quadratic equations is by factoring the equation.
$\,\Rightarrow \text{5}{{\text{y}}^{\text{2}}}\text{+5y-10=0}.....\text{(1)}$
Simplifying equation (1) by taking out 5 as common term and then solving for y we get,
\begin{align} & \,\Rightarrow \,{{\text{y}}^{\text{2}}}\text{+y-2=0} \\ & \,\Rightarrow {{\text{y}}^{\text{2}}}\text{+2y-y-2=0} \\ & \,\Rightarrow \text{y(y+2)-1(y+2)=0} \\ & \,\Rightarrow \text{(y+2) (y-1)=0} \\ & \,\Rightarrow \text{y=}\,-2,\,+1 \\ \end{align}