X^2+10x=567

Solution for X^2+10x=567 equation:

X^2+10X=567

We move all terms to the left:

X^2+10X-(567)=0

a = 1; b = 10; c = -567;
Δ = b2-4ac
Δ = 102-4·1·(-567)
Δ = 2368
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:

$\sqrt{\Delta}=\sqrt{2368}=\sqrt{64*37}=\sqrt{64}*\sqrt{37}=8\sqrt{37}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-8\sqrt{37}}{2*1}=\frac{-10-8\sqrt{37}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+8\sqrt{37}}{2*1}=\frac{-10+8\sqrt{37}}{2}
$