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2.5x^2+40x-90=0
a = 2.5; b = 40; c = -90;
Δ = b2-4ac
Δ = 402-4·2.5·(-90)
Δ = 2500
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{2500}=50$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(40)-50}{2*2.5}=\frac{-90}{5} =-18 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(40)+50}{2*2.5}=\frac{10}{5} =2 $
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