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15x^2+16x-7=0
a = 15; b = 16; c = -7;
Δ = b2-4ac
Δ = 162-4·15·(-7)
Δ = 676
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{676}=26$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-26}{2*15}=\frac{-42}{30} =-1+2/5 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+26}{2*15}=\frac{10}{30} =1/3 $
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