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a2+14a-51=0
We add all the numbers together, and all the variables
a^2+14a-51=0
a = 1; b = 14; c = -51;
Δ = b2-4ac
Δ = 142-4·1·(-51)
Δ = 400
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{400}=20$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-20}{2*1}=\frac{-34}{2} =-17 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+20}{2*1}=\frac{6}{2} =3 $
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