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10x^2-35x+15=0
a = 10; b = -35; c = +15;
Δ = b2-4ac
Δ = -352-4·10·15
Δ = 625
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{625}=25$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-35)-25}{2*10}=\frac{10}{20} =1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-35)+25}{2*10}=\frac{60}{20} =3 $
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