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15x^2+x-16=0
a = 15; b = 1; c = -16;
Δ = b2-4ac
Δ = 12-4·15·(-16)
Δ = 961
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{961}=31$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-31}{2*15}=\frac{-32}{30} =-1+1/15 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+31}{2*15}=\frac{30}{30} =1 $
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