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5x^2+7x-9=4x^2+x-6
We move all terms to the left:
5x^2+7x-9-(4x^2+x-6)=0
We get rid of parentheses
5x^2-4x^2+7x-x+6-9=0
We add all the numbers together, and all the variables
x^2+6x-3=0
a = 1; b = 6; c = -3;
Δ = b2-4ac
Δ = 62-4·1·(-3)
Δ = 48
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{3}}{2*1}=\frac{-6-4\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{3}}{2*1}=\frac{-6+4\sqrt{3}}{2} $
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