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.5x^2+4x-160=0
a = .5; b = 4; c = -160;
Δ = b2-4ac
Δ = 42-4·.5·(-160)
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{21}}{2*.5}=\frac{-4-4\sqrt{21}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{21}}{2*.5}=\frac{-4+4\sqrt{21}}{1} $
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