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2x^2+3x-7=x2+5x+39
We move all terms to the left:
2x^2+3x-7-(x2+5x+39)=0
We add all the numbers together, and all the variables
2x^2-(+x^2+5x+39)+3x-7=0
We get rid of parentheses
2x^2-x^2-5x+3x-39-7=0
We add all the numbers together, and all the variables
x^2-2x-46=0
a = 1; b = -2; c = -46;
Δ = b2-4ac
Δ = -22-4·1·(-46)
Δ = 188
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{188}=\sqrt{4*47}=\sqrt{4}*\sqrt{47}=2\sqrt{47}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{47}}{2*1}=\frac{2-2\sqrt{47}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{47}}{2*1}=\frac{2+2\sqrt{47}}{2} $
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