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.5x^2+16x+3=0
a = .5; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·.5·3
Δ = 250
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{250}=\sqrt{25*10}=\sqrt{25}*\sqrt{10}=5\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-5\sqrt{10}}{2*.5}=\frac{-16-5\sqrt{10}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+5\sqrt{10}}{2*.5}=\frac{-16+5\sqrt{10}}{1} $
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