(3x2-6x+4)+(4x2-8x-5)=

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Solution for (3x2-6x+4)+(4x2-8x-5)= equation:



(3x^2-6x+4)+(4x^2-8x-5)=
We move all terms to the left:
(3x^2-6x+4)+(4x^2-8x-5)-()=0
We add all the numbers together, and all the variables
(3x^2-6x+4)+(4x^2-8x-5)=0
We get rid of parentheses
3x^2+4x^2-6x-8x+4-5=0
We add all the numbers together, and all the variables
7x^2-14x-1=0
a = 7; b = -14; c = -1;
Δ = b2-4ac
Δ = -142-4·7·(-1)
Δ = 224
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:
$\sqrt{\Delta}=\sqrt{224}=\sqrt{16*14}=\sqrt{16}*\sqrt{14}=4\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-4\sqrt{14}}{2*7}=\frac{14-4\sqrt{14}}{14} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+4\sqrt{14}}{2*7}=\frac{14+4\sqrt{14}}{14} $

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