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2a^2+23a+56=0
a = 2; b = 23; c = +56;
Δ = b2-4ac
Δ = 232-4·2·56
Δ = 81
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{81}=9$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(23)-9}{2*2}=\frac{-32}{4} =-8 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(23)+9}{2*2}=\frac{-14}{4} =-3+1/2 $
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