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x^2+12x-10=-4
We move all terms to the left:
x^2+12x-10-(-4)=0
We add all the numbers together, and all the variables
x^2+12x-6=0
a = 1; b = 12; c = -6;
Δ = b2-4ac
Δ = 122-4·1·(-6)
Δ = 168
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{42}}{2*1}=\frac{-12-2\sqrt{42}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{42}}{2*1}=\frac{-12+2\sqrt{42}}{2} $
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