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x^2+2x=40-x
We move all terms to the left:
x^2+2x-(40-x)=0
We add all the numbers together, and all the variables
x^2+2x-(-1x+40)=0
We get rid of parentheses
x^2+2x+1x-40=0
We add all the numbers together, and all the variables
x^2+3x-40=0
a = 1; b = 3; c = -40;
Δ = b2-4ac
Δ = 32-4·1·(-40)
Δ = 169
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{169}=13$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*1}=\frac{-16}{2} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*1}=\frac{10}{2} =5 $
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