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16x^2+21x-17=0
a = 16; b = 21; c = -17;
Δ = b2-4ac
Δ = 212-4·16·(-17)
Δ = 1529
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-\sqrt{1529}}{2*16}=\frac{-21-\sqrt{1529}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+\sqrt{1529}}{2*16}=\frac{-21+\sqrt{1529}}{32} $
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