Question

Solve the given quadratic equation by using the formula method, $2y^2+6y=3$

Answer 1

The given quadratic equation $2y^2+6y=3$ which can be rewritten as$2y^2+6y-3=0$ is in the form $a{x}^2+bx+c=0$ where $a=2,b=6$ and $c=-3$.

We know that the quadratic formula is $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$, therefore, substitute $a=2,b=6$ and $c=-3$ in $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ as follows:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-6\pm \sqrt{6^2-(4\times 2\times -3)}}{2\times 2}=\frac{-6\pm \sqrt{36+24}}{4}=\frac{-6\pm \sqrt{60}}{4}$

$=\frac{-6\pm \sqrt{4\times 15}}{4}=\frac{-6\pm 2\sqrt{15}}{4}=\frac{2(-3\pm \sqrt{15})}{4}=\frac{-3\pm \sqrt{15}}{2}$

Hence, $x=\frac{-3+\sqrt{15}}{2}$ or $x=\frac{-3-\sqrt{15}}{2}$.