x^2+21=10x-3

Solution for x^2+21=10x-3 equation:

x^2+21=10x-3

We move all terms to the left:

x^2+21-(10x-3)=0

We get rid of parentheses

x^2-10x+3+21=0

We add all the numbers together, and all the variables

x^2-10x+24=0

a = 1; b = -10; c = +24;
Δ = b2-4ac
Δ = -102-4·1·24
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2}{2*1}=\frac{8}{2}
=4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2}{2*1}=\frac{12}{2}
=6
$