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2x^2+8x=5x+5
We move all terms to the left:
2x^2+8x-(5x+5)=0
We get rid of parentheses
2x^2+8x-5x-5=0
We add all the numbers together, and all the variables
2x^2+3x-5=0
a = 2; b = 3; c = -5;
Δ = b2-4ac
Δ = 32-4·2·(-5)
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-7}{2*2}=\frac{-10}{4} =-2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+7}{2*2}=\frac{4}{4} =1 $
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