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15x^2-16x+1=0
a = 15; b = -16; c = +1;
Δ = b2-4ac
Δ = -162-4·15·1
Δ = 196
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{196}=14$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-14}{2*15}=\frac{2}{30} =1/15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+14}{2*15}=\frac{30}{30} =1 $
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