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4x^2+19x-140=0
a = 4; b = 19; c = -140;
Δ = b2-4ac
Δ = 192-4·4·(-140)
Δ = 2601
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2601}=51$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-51}{2*4}=\frac{-70}{8} =-8+3/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+51}{2*4}=\frac{32}{8} =4 $
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