Question

Solve the equation using quadratic formula.

$2x^2+12x-3=0$

Short answer

The roots are$x=\frac{-6+\sqrt{42}}{2},x=\frac{-6-\sqrt{42}}{2}$.

Step-by-step solution

 Given Information.     

The given equation is$2x^2+12x-3=0$.

Write the quadratic equation in standard form and identify  a,b,c.

$2x^2+12x-3=0$

The equation is in standard form.

So,$a=2,b=12,c-3$.

Substitute the values in quadratic formula.     

$$\begin{gathered} x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ x=\frac{-12\pm \sqrt{(12{)}^2-4(2\left)\right(-3)}}{2\left(2\right)} \\ x=\frac{-12\pm \sqrt{144+24}}{4} \\ x=\frac{-12\pm \sqrt{168}}{4} \end{gathered}$$

Use square root property.  

$$\begin{gathered} x=\frac{-12\pm 4\sqrt{42}}{4} \\ x=\frac{2(-6\pm \sqrt{42})}{4} \\ x=\frac{-6\pm \sqrt{42}}{2} \end{gathered}$$

Rewrite to show two solutions.

$$\begin{gathered} x=\frac{-6+\sqrt{42}}{2}, \\ x=\frac{-6-\sqrt{42}}{2} \end{gathered}$$