-3x(x-6)+4(x+1)=7x-8

Solution for -3x(x-6)+4(x+1)=7x-8 equation:

-3x(x-6)+4(x+1)=7x-8

We move all terms to the left:

-3x(x-6)+4(x+1)-(7x-8)=0

We multiply parentheses

-3x^2+18x+4x-(7x-8)+4=0

We get rid of parentheses

-3x^2+18x+4x-7x+8+4=0

We add all the numbers together, and all the variables

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

a = -3; b = 15; c = +12;
Δ = b2-4ac
Δ = 152-4·(-3)·12
Δ = 369
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:

$\sqrt{\Delta}=\sqrt{369}=\sqrt{9*41}=\sqrt{9}*\sqrt{41}=3\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{41}}{2*-3}=\frac{-15-3\sqrt{41}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{41}}{2*-3}=\frac{-15+3\sqrt{41}}{-6}
$