3x^2+20x=2250

Solution for 3x^2+20x=2250 equation:

3x^2+20x=2250

We move all terms to the left:

3x^2+20x-(2250)=0

a = 3; b = 20; c = -2250;
Δ = b2-4ac
Δ = 202-4·3·(-2250)
Δ = 27400
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:

$\sqrt{\Delta}=\sqrt{27400}=\sqrt{100*274}=\sqrt{100}*\sqrt{274}=10\sqrt{274}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10\sqrt{274}}{2*3}=\frac{-20-10\sqrt{274}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10\sqrt{274}}{2*3}=\frac{-20+10\sqrt{274}}{6}
$