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4x^2+31x-45=0
a = 4; b = 31; c = -45;
Δ = b2-4ac
Δ = 312-4·4·(-45)
Δ = 1681
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{1681}=41$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(31)-41}{2*4}=\frac{-72}{8} =-9 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(31)+41}{2*4}=\frac{10}{8} =1+1/4 $
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