Answer:
[tex]x=\dfrac{2}{3},-\dfrac{3}{2}[/tex]
Step-by-step explanation:
Given equation:
[tex]6x^2+5x-6=0[/tex]
First, factor the left side of the given equation.
To factor a quadratic in the form [tex]ax^2+bx+c[/tex] find two numbers that multiply to [tex]ac[/tex] and sum to [tex]b[/tex]:
[tex]\implies ac=6\cdot-6=-36[/tex]
[tex]\implies b=5[/tex]
So the two numbers are: 9 and -4
Rewrite [tex]b[/tex] as the sum of these two numbers:
[tex]\implies 6x^2+9x-4x-6=0[/tex]
Factorize the first two terms and the last two terms separately:
[tex]\implies 3x(2x+3)-2(2x+3)=0[/tex]
Factor out the common term [tex](2x+3)[/tex]:
[tex]\implies (3x-2)(2x+3)=0[/tex]
To solve for x:
[tex]\begin{aligned}\implies (3x-2) & =0 & \implies (2x+3) & = 0\\3x & = 2 & 2x & = -3\\x & = \dfrac{2}{3} & x & = -\dfrac{3}{2}\end{aligned}[/tex]
Therefore:
[tex]x=\dfrac{2}{3},-\dfrac{3}{2}[/tex]