2x^2=5x(x-7)+4

Solution for 2x^2=5x(x-7)+4 equation:

2x^2=5x(x-7)+4

We move all terms to the left:

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


We calculate terms in parentheses: -(5x(x-7)+4), so:

5x(x-7)+4

We multiply parentheses

5x^2-35x+4

Back to the equation:

-(5x^2-35x+4)


We get rid of parentheses

2x^2-5x^2+35x-4=0

We add all the numbers together, and all the variables

-3x^2+35x-4=0

a = -3; b = 35; c = -4;
Δ = b2-4ac
Δ = 352-4·(-3)·(-4)
Δ = 1177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(35)-\sqrt{1177}}{2*-3}=\frac{-35-\sqrt{1177}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(35)+\sqrt{1177}}{2*-3}=\frac{-35+\sqrt{1177}}{-6}
$