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a(a-3)-40=0
We multiply parentheses
a^2-3a-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:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{169}=13$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-13}{2*1}=\frac{-10}{2} =-5 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+13}{2*1}=\frac{16}{2} =8 $
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