3x^2-4x+5=20

Solution for 3x^2-4x+5=20 equation:

3x^2-4x+5=20

We move all terms to the left:

3x^2-4x+5-(20)=0

We add all the numbers together, and all the variables

3x^2-4x-15=0

a = 3; b = -4; c = -15;
Δ = b2-4ac
Δ = -42-4·3·(-15)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14}{2*3}=\frac{-10}{6}
=-1+2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14}{2*3}=\frac{18}{6}
=3
$