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5x^2+43x+24=0
a = 5; b = 43; c = +24;
Δ = b2-4ac
Δ = 432-4·5·24
Δ = 1369
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{1369}=37$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(43)-37}{2*5}=\frac{-80}{10} =-8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(43)+37}{2*5}=\frac{-6}{10} =-3/5 $
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