x^2+3(x-6)-7x+12,x=3

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Solution for x^2+3(x-6)-7x+12,x=3 equation:



x^2+3(x-6)-7x+12.x=3
We move all terms to the left:
x^2+3(x-6)-7x+12.x-(3)=0
We add all the numbers together, and all the variables
x^2+5x+3(x-6)-3=0
We multiply parentheses
x^2+5x+3x-18-3=0
We add all the numbers together, and all the variables
x^2+8x-21=0
a = 1; b = 8; c = -21;
Δ = b2-4ac
Δ = 82-4·1·(-21)
Δ = 148
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:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{37}}{2*1}=\frac{-8-2\sqrt{37}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{37}}{2*1}=\frac{-8+2\sqrt{37}}{2} $

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