16x^2-40x+25x=0

Solution for 16x^2-40x+25x=0 equation:

16x^2-40x+25x=0

We add all the numbers together, and all the variables

16x^2-15x=0

a = 16; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·16·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*16}=\frac{0}{32}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*16}=\frac{30}{32}
=15/16
$