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2x^2-5x+4=3x+8
We move all terms to the left:
2x^2-5x+4-(3x+8)=0
We get rid of parentheses
2x^2-5x-3x-8+4=0
We add all the numbers together, and all the variables
2x^2-8x-4=0
a = 2; b = -8; c = -4;
Δ = b2-4ac
Δ = -82-4·2·(-4)
Δ = 96
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{96}=\sqrt{16*6}=\sqrt{16}*\sqrt{6}=4\sqrt{6}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-4\sqrt{6}}{2*2}=\frac{8-4\sqrt{6}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+4\sqrt{6}}{2*2}=\frac{8+4\sqrt{6}}{4} $
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