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16x^2+28x-35=0
a = 16; b = 28; c = -35;
Δ = b2-4ac
Δ = 282-4·16·(-35)
Δ = 3024
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3024}=\sqrt{144*21}=\sqrt{144}*\sqrt{21}=12\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(28)-12\sqrt{21}}{2*16}=\frac{-28-12\sqrt{21}}{32} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(28)+12\sqrt{21}}{2*16}=\frac{-28+12\sqrt{21}}{32} $
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