2x^2+10x=18x+42

Solution for 2x^2+10x=18x+42 equation:

2x^2+10x=18x+42

We move all terms to the left:

2x^2+10x-(18x+42)=0

We get rid of parentheses

2x^2+10x-18x-42=0

We add all the numbers together, and all the variables

2x^2-8x-42=0

a = 2; b = -8; c = -42;
Δ = b2-4ac
Δ = -82-4·2·(-42)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-20}{2*2}=\frac{-12}{4}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+20}{2*2}=\frac{28}{4}
=7
$