d2-19d+90=0

Solution for d2-19d+90=0 equation:

d2-19d+90=0

We add all the numbers together, and all the variables

d^2-19d+90=0

a = 1; b = -19; c = +90;
Δ = b2-4ac
Δ = -192-4·1·90
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-1}{2*1}=\frac{18}{2}
=9
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+1}{2*1}=\frac{20}{2}
=10
$