X^2+10x=2400

Solution for X^2+10x=2400 equation:

X^2+10X=2400

We move all terms to the left:

X^2+10X-(2400)=0

a = 1; b = 10; c = -2400;
Δ = b2-4ac
Δ = 102-4·1·(-2400)
Δ = 9700
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:

$\sqrt{\Delta}=\sqrt{9700}=\sqrt{100*97}=\sqrt{100}*\sqrt{97}=10\sqrt{97}$
$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10\sqrt{97}}{2*1}=\frac{-10-10\sqrt{97}}{2}
$
$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10\sqrt{97}}{2*1}=\frac{-10+10\sqrt{97}}{2}
$