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2x^2-15+x=0
a = 2; b = 1; c = -15;
Δ = b2-4ac
Δ = 12-4·2·(-15)
Δ = 121
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{121}=11$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-11}{2*2}=\frac{-12}{4} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+11}{2*2}=\frac{10}{4} =2+1/2 $
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