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14a^2-19a-40=0
a = 14; b = -19; c = -40;
Δ = b2-4ac
Δ = -192-4·14·(-40)
Δ = 2601
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{2601}=51$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-51}{2*14}=\frac{-32}{28} =-1+1/7 $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+51}{2*14}=\frac{70}{28} =2+1/2 $
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