4^2+10x=x^2-x+4

Solution for 4^2+10x=x^2-x+4 equation:

4^2+10x=x^2-x+4

We move all terms to the left:

4^2+10x-(x^2-x+4)=0

We add all the numbers together, and all the variables

10x-(x^2-x+4)+16=0

We get rid of parentheses

-x^2+10x+x-4+16=0

We add all the numbers together, and all the variables

-1x^2+11x+12=0

a = -1; b = 11; c = +12;
Δ = b2-4ac
Δ = 112-4·(-1)·12
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-13}{2*-1}=\frac{-24}{-2}
=+12
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+13}{2*-1}=\frac{2}{-2}
=-1
$