91Ó°ÊÓ

The given equation is

$2b^2+6b−8=0$

First, write the equation $a{x}^2+bx+c=0$in standard form, as shown below.

This is already done in this place.

Determine the a, b, and c values.

When compared to the standard, the values are a = 2 b = 6 and c= -8.

In the quadratic formula, substitute a = 2,b = 6, and c = -8.

$x=\frac{-bÂ\pm \sqrt{b^2-4ac}}{2a}$

localid="1653828197551" $b=\frac{-6Â\pm \sqrt{{\left(6\right)}^2-4\left(2\right)(-8)}}{2\left(2\right)}$

As shown below, simplify the fraction.

$$\begin{gathered} b=\frac{-6Â\pm \sqrt{{\left(6\right)}^2-4\left(2\right)(-8)}}{2\left(2\right)} \\ b=\frac{-6Â\pm \sqrt{36+64}}{4} \\ b=\frac{-6+\sqrt{100}}{4} \\ b=\frac{-6Â\pm 10}{4} \\ 1,-4 \end{gathered}$$

As a result, the two solutions are b = 1 and b = -4.

To test the solutions, enter b = 1 and b = -4 into the original equation.

For b = 1

localid="1653828712202" $$\begin{gathered} 2b^2+6b-8=0 \\ 2b^2+6b-8-0 \\ 2{\left(1\right)}^2+6\left(1\right)-8=0 \\ 2+6-8=0 \\ 0=0 \end{gathered}$$

For b = -4

localid="1657799462454" $$\begin{gathered} 2b^2+6b-8=0 \\ 2b^2+6b-8-0 \\ 2{(-4)}^2+6(-4)-8=0 \\ 32-24-8=0 \\ 0=0 \end{gathered}$$

As a result, the solutions are b = 1 and b = -4.