7x+15=3x^2-19x+12

Solution for 7x+15=3x^2-19x+12 equation:

7x+15=3x^2-19x+12

We move all terms to the left:

7x+15-(3x^2-19x+12)=0

We get rid of parentheses

-3x^2+7x+19x-12+15=0

We add all the numbers together, and all the variables

-3x^2+26x+3=0

a = -3; b = 26; c = +3;
Δ = b2-4ac
Δ = 262-4·(-3)·3
Δ = 712
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{712}=\sqrt{4*178}=\sqrt{4}*\sqrt{178}=2\sqrt{178}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-2\sqrt{178}}{2*-3}=\frac{-26-2\sqrt{178}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+2\sqrt{178}}{2*-3}=\frac{-26+2\sqrt{178}}{-6}
$