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-12x^2+10x+590=0
a = -12; b = 10; c = +590;
Δ = b2-4ac
Δ = 102-4·(-12)·590
Δ = 28420
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{28420}=\sqrt{196*145}=\sqrt{196}*\sqrt{145}=14\sqrt{145}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14\sqrt{145}}{2*-12}=\frac{-10-14\sqrt{145}}{-24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14\sqrt{145}}{2*-12}=\frac{-10+14\sqrt{145}}{-24} $
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