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2x^2+27x-62=0
a = 2; b = 27; c = -62;
Δ = b2-4ac
Δ = 272-4·2·(-62)
Δ = 1225
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{1225}=35$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-35}{2*2}=\frac{-62}{4} =-15+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+35}{2*2}=\frac{8}{4} =2 $
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