4x^2+300x+1495=0

Solution for 4x^2+300x+1495=0 equation:

4x^2+300x+1495=0

a = 4; b = 300; c = +1495;
Δ = b2-4ac
Δ = 3002-4·4·1495
Δ = 66080
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:

$\sqrt{\Delta}=\sqrt{66080}=\sqrt{16*4130}=\sqrt{16}*\sqrt{4130}=4\sqrt{4130}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(300)-4\sqrt{4130}}{2*4}=\frac{-300-4\sqrt{4130}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(300)+4\sqrt{4130}}{2*4}=\frac{-300+4\sqrt{4130}}{8}
$