3x^2+8x+10=65

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Solution for 3x^2+8x+10=65 equation:



3x^2+8x+10=65
We move all terms to the left:
3x^2+8x+10-(65)=0
We add all the numbers together, and all the variables
3x^2+8x-55=0
a = 3; b = 8; c = -55;
Δ = b2-4ac
Δ = 82-4·3·(-55)
Δ = 724
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:
$\sqrt{\Delta}=\sqrt{724}=\sqrt{4*181}=\sqrt{4}*\sqrt{181}=2\sqrt{181}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{181}}{2*3}=\frac{-8-2\sqrt{181}}{6} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{181}}{2*3}=\frac{-8+2\sqrt{181}}{6} $

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