2x^2+21x=220

Solution for 2x^2+21x=220 equation:

2x^2+21x=220

We move all terms to the left:

2x^2+21x-(220)=0

a = 2; b = 21; c = -220;
Δ = b2-4ac
Δ = 212-4·2·(-220)
Δ = 2201
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{2201}}{2*2}=\frac{-21-\sqrt{2201}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{2201}}{2*2}=\frac{-21+\sqrt{2201}}{4}
$