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.5x^2+4x-600=0
a = .5; b = 4; c = -600;
Δ = b2-4ac
Δ = 42-4·.5·(-600)
Δ = 1216
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1216}=\sqrt{64*19}=\sqrt{64}*\sqrt{19}=8\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-8\sqrt{19}}{2*.5}=\frac{-4-8\sqrt{19}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+8\sqrt{19}}{2*.5}=\frac{-4+8\sqrt{19}}{1} $
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