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2x^2+2x=5x+35
We move all terms to the left:
2x^2+2x-(5x+35)=0
We get rid of parentheses
2x^2+2x-5x-35=0
We add all the numbers together, and all the variables
2x^2-3x-35=0
a = 2; b = -3; c = -35;
Δ = b2-4ac
Δ = -32-4·2·(-35)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-17}{2*2}=\frac{-14}{4} =-3+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+17}{2*2}=\frac{20}{4} =5 $
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