-x^2+5x+16=x+6

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Solution for -x^2+5x+16=x+6 equation:



-x^2+5x+16=x+6
We move all terms to the left:
-x^2+5x+16-(x+6)=0
We add all the numbers together, and all the variables
-1x^2+5x-(x+6)+16=0
We get rid of parentheses
-1x^2+5x-x-6+16=0
We add all the numbers together, and all the variables
-1x^2+4x+10=0
a = -1; b = 4; c = +10;
Δ = b2-4ac
Δ = 42-4·(-1)·10
Δ = 56
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:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{14}}{2*-1}=\frac{-4-2\sqrt{14}}{-2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{14}}{2*-1}=\frac{-4+2\sqrt{14}}{-2} $

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