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X2+6X-3=0
We add all the numbers together, and all the variables
X^2+6X-3=0
a = 1; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·1·(-3)
Δ = 48
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}$$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*1}=\frac{-6-4\sqrt{3}}{2} $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*1}=\frac{-6+4\sqrt{3}}{2} $
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