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4x+3=x^2-6+5
We move all terms to the left:
4x+3-(x^2-6+5)=0
We get rid of parentheses
-x^2+4x+6-5+3=0
We add all the numbers together, and all the variables
-1x^2+4x+4=0
a = -1; b = 4; c = +4;
Δ = b2-4ac
Δ = 42-4·(-1)·4
Δ = 32
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{2}}{2*-1}=\frac{-4-4\sqrt{2}}{-2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{2}}{2*-1}=\frac{-4+4\sqrt{2}}{-2} $
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