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2x^2+3x-5=7x-6
We move all terms to the left:
2x^2+3x-5-(7x-6)=0
We get rid of parentheses
2x^2+3x-7x+6-5=0
We add all the numbers together, and all the variables
2x^2-4x+1=0
a = 2; b = -4; c = +1;
Δ = b2-4ac
Δ = -42-4·2·1
Δ = 8
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8}=\sqrt{4*2}=\sqrt{4}*\sqrt{2}=2\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{2}}{2*2}=\frac{4-2\sqrt{2}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{2}}{2*2}=\frac{4+2\sqrt{2}}{4} $
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