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2x^2-15x+25=0
a = 2; b = -15; c = +25;
Δ = b2-4ac
Δ = -152-4·2·25
Δ = 25
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{25}=5$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-5}{2*2}=\frac{10}{4} =2+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+5}{2*2}=\frac{20}{4} =5 $
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