2b^2-18b+42=6

Solution for 2b^2-18b+42=6 equation:

2b^2-18b+42=6

We move all terms to the left:

2b^2-18b+42-(6)=0

We add all the numbers together, and all the variables

2b^2-18b+36=0

a = 2; b = -18; c = +36;
Δ = b2-4ac
Δ = -182-4·2·36
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6}{2*2}=\frac{12}{4}
=3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6}{2*2}=\frac{24}{4}
=6
$