2x^2=x(x-14)-5

Solution for 2x^2=x(x-14)-5 equation:

2x^2=x(x-14)-5

We move all terms to the left:

2x^2-(x(x-14)-5)=0


We calculate terms in parentheses: -(x(x-14)-5), so:

x(x-14)-5

We multiply parentheses

x^2-14x-5

Back to the equation:

-(x^2-14x-5)


We get rid of parentheses

2x^2-x^2+14x+5=0

We add all the numbers together, and all the variables

x^2+14x+5=0

a = 1; b = 14; c = +5;
Δ = b2-4ac
Δ = 142-4·1·5
Δ = 176
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:

$\sqrt{\Delta}=\sqrt{176}=\sqrt{16*11}=\sqrt{16}*\sqrt{11}=4\sqrt{11}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-4\sqrt{11}}{2*1}=\frac{-14-4\sqrt{11}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+4\sqrt{11}}{2*1}=\frac{-14+4\sqrt{11}}{2}
$