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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

-1x^2+14x+4(9x-5)/2-21=0

We multiply all the terms by the denominator


-1x^2*2+14x*2+4(9x-5)-21*2=0

We add all the numbers together, and all the variables

-1x^2*2+14x*2+4(9x-5)-42=0

We multiply parentheses

-1x^2*2+14x*2+36x-20-42=0

Wy multiply elements

-2x^2+28x+36x-20-42=0

We add all the numbers together, and all the variables

-2x^2+64x-62=0

a = -2; b = 64; c = -62;
Δ = b2-4ac
Δ = 642-4·(-2)·(-62)
Δ = 3600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{3600}=60$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(64)-60}{2*-2}=\frac{-124}{-4}
=+31
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(64)+60}{2*-2}=\frac{-4}{-4}
=1
$