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

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

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

We move all terms to the left:

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

We multiply parentheses

-3x^2+18x+(4x+1)-(7x-8)=0

We get rid of parentheses

-3x^2+18x+4x-7x+1+8=0

We add all the numbers together, and all the variables

-3x^2+15x+9=0

a = -3; b = 15; c = +9;
Δ = b2-4ac
Δ = 152-4·(-3)·9
Δ = 333
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:

$\sqrt{\Delta}=\sqrt{333}=\sqrt{9*37}=\sqrt{9}*\sqrt{37}=3\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-3\sqrt{37}}{2*-3}=\frac{-15-3\sqrt{37}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+3\sqrt{37}}{2*-3}=\frac{-15+3\sqrt{37}}{-6}
$