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3x^2+2x-20=2x^2+15
We move all terms to the left:
3x^2+2x-20-(2x^2+15)=0
We get rid of parentheses
3x^2-2x^2+2x-15-20=0
We add all the numbers together, and all the variables
x^2+2x-35=0
a = 1; b = 2; c = -35;
Δ = b2-4ac
Δ = 22-4·1·(-35)
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-12}{2*1}=\frac{-14}{2} =-7 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+12}{2*1}=\frac{10}{2} =5 $
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