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15x^2-10-5x=0
a = 15; b = -5; c = -10;
Δ = b2-4ac
Δ = -52-4·15·(-10)
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-25}{2*15}=\frac{-20}{30} =-2/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+25}{2*15}=\frac{30}{30} =1 $
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