Evaluate: 8 x ^ 2+2 x-3 = 0

Expression: $$8 x ^ { 2 } + 2 x - 3 = 0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $8x^{2}+ax+bx-3$. To find $a$ and $b$, set up a system to be solved.

$$a+b=2$$ $$ab=8\left(-3\right)=-24$$

Since $ab$ is negative, $a$ and $b$ have the opposite signs. Since $a+b$ is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product $-24$.

$$-1,24$$ $$-2,12$$ $$-3,8$$ $$-4,6$$

Calculate the sum for each pair.

$$-1+24=23$$ $$-2+12=10$$ $$-3+8=5$$ $$-4+6=2$$

The solution is the pair that gives sum $2$.

$$a=-4$$ $$b=6$$

Rewrite $8x^{2}+2x-3$ as $\left(8x^{2}-4x\right)+\left(6x-3\right)$.

$$\left(8x^{2}-4x\right)+\left(6x-3\right)$$

Factor out $4x$ in the first and $3$ in the second group.

$$4x\left(2x-1\right)+3\left(2x-1\right)$$

Factor out common term $2x-1$ by using distributive property.

$$\left(2x-1\right)\left(4x+3\right)$$

To find equation solutions, solve $2x-1=0$ and $4x+3=0$.

$$x=\frac{1}{2}$$ $$x=-\frac{3}{4}$$