7(2x-3)+4(9x-5)10=x2

Solution for 7(2x-3)+4(9x-5)10=x2 equation:

7(2x-3)+4(9x-5)10=x2

We move all terms to the left:

7(2x-3)+4(9x-5)10-(x2)=0

We add all the numbers together, and all the variables

-1x^2+7(2x-3)+4(9x-5)10=0

We multiply parentheses

-1x^2+14x+360x-21-200=0

We add all the numbers together, and all the variables

-1x^2+374x-221=0

a = -1; b = 374; c = -221;
Δ = b2-4ac
Δ = 3742-4·(-1)·(-221)
Δ = 138992
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:

$\sqrt{\Delta}=\sqrt{138992}=\sqrt{16*8687}=\sqrt{16}*\sqrt{8687}=4\sqrt{8687}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(374)-4\sqrt{8687}}{2*-1}=\frac{-374-4\sqrt{8687}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(374)+4\sqrt{8687}}{2*-1}=\frac{-374+4\sqrt{8687}}{-2}
$