32x^2+65x+25=0

Solution for 32x^2+65x+25=0 equation:

32x^2+65x+25=0

a = 32; b = 65; c = +25;
Δ = b2-4ac
Δ = 652-4·32·25
Δ = 1025
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:

$\sqrt{\Delta}=\sqrt{1025}=\sqrt{25*41}=\sqrt{25}*\sqrt{41}=5\sqrt{41}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(65)-5\sqrt{41}}{2*32}=\frac{-65-5\sqrt{41}}{64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(65)+5\sqrt{41}}{2*32}=\frac{-65+5\sqrt{41}}{64}
$