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0.5x^2-10x+12=0
a = 0.5; b = -10; c = +12;
Δ = b2-4ac
Δ = -102-4·0.5·12
Δ = 76
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{19}}{2*0.5}=\frac{10-2\sqrt{19}}{1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{19}}{2*0.5}=\frac{10+2\sqrt{19}}{1} $
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