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4x^2+2x-151=0
a = 4; b = 2; c = -151;
Δ = b2-4ac
Δ = 22-4·4·(-151)
Δ = 2420
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2420}=\sqrt{484*5}=\sqrt{484}*\sqrt{5}=22\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-22\sqrt{5}}{2*4}=\frac{-2-22\sqrt{5}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+22\sqrt{5}}{2*4}=\frac{-2+22\sqrt{5}}{8} $
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