32x-16x^2-15=0

Solution for 32x-16x^2-15=0 equation:

32x-16x^2-15=0

a = -16; b = 32; c = -15;
Δ = b2-4ac
Δ = 322-4·(-16)·(-15)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-8}{2*-16}=\frac{-40}{-32}
=1+1/4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+8}{2*-16}=\frac{-24}{-32}
=3/4
$