(3x-1)(x+2)=10

Solution for (3x-1)(x+2)=10 equation:

(3x-1)(x+2)=10

We move all terms to the left:

(3x-1)(x+2)-(10)=0

We multiply parentheses ..

(+3x^2+6x-1x-2)-10=0

We get rid of parentheses

3x^2+6x-1x-2-10=0

We add all the numbers together, and all the variables

3x^2+5x-12=0

a = 3; b = 5; c = -12;
Δ = b2-4ac
Δ = 52-4·3·(-12)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(5)-13}{2*3}=\frac{-18}{6}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+13}{2*3}=\frac{8}{6}
=1+1/3
$