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X2+13X-40=8
We move all terms to the left:
X2+13X-40-(8)=0
We add all the numbers together, and all the variables
X^2+13X-48=0
a = 1; b = 13; c = -48;
Δ = b2-4ac
Δ = 132-4·1·(-48)
Δ = 361
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{361}=19$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(13)-19}{2*1}=\frac{-32}{2} =-16 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+19}{2*1}=\frac{6}{2} =3 $
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