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.5x^2+55x-15=0
a = .5; b = 55; c = -15;
Δ = b2-4ac
Δ = 552-4·.5·(-15)
Δ = 3055
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(55)-\sqrt{3055}}{2*.5}=\frac{-55-\sqrt{3055}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(55)+\sqrt{3055}}{2*.5}=\frac{-55+\sqrt{3055}}{1} $
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