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x^2+6x-3=-2
We move all terms to the left:
x^2+6x-3-(-2)=0
We add all the numbers together, and all the variables
x^2+6x-1=0
a = 1; b = 6; c = -1;
Δ = b2-4ac
Δ = 62-4·1·(-1)
Δ = 40
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{10}}{2*1}=\frac{-6-2\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{10}}{2*1}=\frac{-6+2\sqrt{10}}{2} $
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