5x^2-20x+19=4

Solution for 5x^2-20x+19=4 equation:

5x^2-20x+19=4

We move all terms to the left:

5x^2-20x+19-(4)=0

We add all the numbers together, and all the variables

5x^2-20x+15=0

a = 5; b = -20; c = +15;
Δ = b2-4ac
Δ = -202-4·5·15
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10}{2*5}=\frac{10}{10}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10}{2*5}=\frac{30}{10}
=3
$