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12x^2+12x-340=0
a = 12; b = 12; c = -340;
Δ = b2-4ac
Δ = 122-4·12·(-340)
Δ = 16464
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{16464}=\sqrt{784*21}=\sqrt{784}*\sqrt{21}=28\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-28\sqrt{21}}{2*12}=\frac{-12-28\sqrt{21}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+28\sqrt{21}}{2*12}=\frac{-12+28\sqrt{21}}{24} $
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