3x^2+16x-2250=0

Solution for 3x^2+16x-2250=0 equation:

3x^2+16x-2250=0

a = 3; b = 16; c = -2250;
Δ = b2-4ac
Δ = 162-4·3·(-2250)
Δ = 27256
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:

$\sqrt{\Delta}=\sqrt{27256}=\sqrt{4*6814}=\sqrt{4}*\sqrt{6814}=2\sqrt{6814}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-2\sqrt{6814}}{2*3}=\frac{-16-2\sqrt{6814}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+2\sqrt{6814}}{2*3}=\frac{-16+2\sqrt{6814}}{6}
$