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x2+8x=40
We move all terms to the left:
x2+8x-(40)=0
We add all the numbers together, and all the variables
x^2+8x-40=0
a = 1; b = 8; c = -40;
Δ = b2-4ac
Δ = 82-4·1·(-40)
Δ = 224
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-4\sqrt{14}}{2*1}=\frac{-8-4\sqrt{14}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+4\sqrt{14}}{2*1}=\frac{-8+4\sqrt{14}}{2} $
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