7(2x-3)+4(9x-5)=x2

Solution for 7(2x-3)+4(9x-5)=x2 equation:

7(2x-3)+4(9x-5)=x2

We move all terms to the left:

7(2x-3)+4(9x-5)-(x2)=0

We add all the numbers together, and all the variables

-1x^2+7(2x-3)+4(9x-5)=0

We multiply parentheses

-1x^2+14x+36x-21-20=0

We add all the numbers together, and all the variables

-1x^2+50x-41=0

a = -1; b = 50; c = -41;
Δ = b2-4ac
Δ = 502-4·(-1)·(-41)
Δ = 2336
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:

$\sqrt{\Delta}=\sqrt{2336}=\sqrt{16*146}=\sqrt{16}*\sqrt{146}=4\sqrt{146}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(50)-4\sqrt{146}}{2*-1}=\frac{-50-4\sqrt{146}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(50)+4\sqrt{146}}{2*-1}=\frac{-50+4\sqrt{146}}{-2}
$