4x^2+100x=7500

Solution for 4x^2+100x=7500 equation:

4x^2+100x=7500

We move all terms to the left:

4x^2+100x-(7500)=0

a = 4; b = 100; c = -7500;
Δ = b2-4ac
Δ = 1002-4·4·(-7500)
Δ = 130000
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:

$\sqrt{\Delta}=\sqrt{130000}=\sqrt{10000*13}=\sqrt{10000}*\sqrt{13}=100\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(100)-100\sqrt{13}}{2*4}=\frac{-100-100\sqrt{13}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(100)+100\sqrt{13}}{2*4}=\frac{-100+100\sqrt{13}}{8}
$