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(X-3)X=40
We move all terms to the left:
(X-3)X-(40)=0
We multiply parentheses
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$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*1}=\frac{-10}{2} =-5 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*1}=\frac{16}{2} =8 $
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