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3x^2+18x-26=+
We move all terms to the left:
3x^2+18x-26-(+)=0
We add all the numbers together, and all the variables
3x^2+18x-26-0=0
We add all the numbers together, and all the variables
3x^2+18x-26=0
a = 3; b = 18; c = -26;
Δ = b2-4ac
Δ = 182-4·3·(-26)
Δ = 636
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{636}=\sqrt{4*159}=\sqrt{4}*\sqrt{159}=2\sqrt{159}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{159}}{2*3}=\frac{-18-2\sqrt{159}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{159}}{2*3}=\frac{-18+2\sqrt{159}}{6} $
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