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2x^2=3-8x
We move all terms to the left:
2x^2-(3-8x)=0
We add all the numbers together, and all the variables
2x^2-(-8x+3)=0
We get rid of parentheses
2x^2+8x-3=0
a = 2; b = 8; c = -3;
Δ = b2-4ac
Δ = 82-4·2·(-3)
Δ = 88
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{22}}{2*2}=\frac{-8-2\sqrt{22}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{22}}{2*2}=\frac{-8+2\sqrt{22}}{4} $
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