Answer
The solutions are - $\frac{4}{3}$ and $\frac{3}{4}$.
Work Step by Step
12$x^{2}$+7x-12=0
12$x^{2}$+16x-9x-12=0
4x(3x+4)-3(3x+4)=0
(4x-3)(3x+4)=0
4x-3 =0 or 3x+4=0
x=$\frac{3}{4}$ 0r x=-$\frac{4}{3}$
The solutions are - $\frac{4}{3}$ and $\frac{3}{4}$.
Check
Let x=- $\frac{4}{3}$
12$x^{2}$+7x-12=0
12$(-4/3)^{2}$+7$\frac{-4}{3}$-12=0
12*16/9-$\frac{28}{3}$-12=0
4*16/3-$\frac{28}{3}$-2=0
64/3-$\frac{28}{3}$-12=0
$\frac{36}{3}$-12=0
12-12=0
0=0
Let x= $\frac{3}{4}$
12$x^{2}$+7x-12=0
12$(3/4)^{2}$+7$\frac{3}{4}$-12=0
12*9/16+$\frac{21}{4}$-12=0
108/16+$\frac{21}{4}$-12=0
27/4+$\frac{21}{4}$-12=0
$\frac{48}{4}$-12=0
12-12=0
0=0
