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12x^2-160x+300=0
a = 12; b = -160; c = +300;
Δ = b2-4ac
Δ = -1602-4·12·300
Δ = 11200
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{11200}=\sqrt{1600*7}=\sqrt{1600}*\sqrt{7}=40\sqrt{7}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-160)-40\sqrt{7}}{2*12}=\frac{160-40\sqrt{7}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-160)+40\sqrt{7}}{2*12}=\frac{160+40\sqrt{7}}{24} $
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