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2x^2-15x-8=0
a = 2; b = -15; c = -8;
Δ = b2-4ac
Δ = -152-4·2·(-8)
Δ = 289
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{289}=17$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-17}{2*2}=\frac{-2}{4} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+17}{2*2}=\frac{32}{4} =8 $
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