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a2+4a=60
We move all terms to the left:
a2+4a-(60)=0
We add all the numbers together, and all the variables
a^2+4a-60=0
a = 1; b = 4; c = -60;
Δ = b2-4ac
Δ = 42-4·1·(-60)
Δ = 256
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{256}=16$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*1}=\frac{-20}{2} =-10 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*1}=\frac{12}{2} =6 $
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