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-0.5x^2-40x+350=0
a = -0.5; b = -40; c = +350;
Δ = b2-4ac
Δ = -402-4·(-0.5)·350
Δ = 2300
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{23}}{2*-0.5}=\frac{40-10\sqrt{23}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{23}}{2*-0.5}=\frac{40+10\sqrt{23}}{-1} $
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