4b^2-25b+36=0

Solution for 4b^2-25b+36=0 equation:

4b^2-25b+36=0

a = 4; b = -25; c = +36;
Δ = b2-4ac
Δ = -252-4·4·36
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-7}{2*4}=\frac{18}{8}
=2+1/4
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+7}{2*4}=\frac{32}{8}
=4
$