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4(2x^2+5)=3(2x^2+8)
We move all terms to the left:
4(2x^2+5)-(3(2x^2+8))=0
We multiply parentheses
8x^2-(3(2x^2+8))+20=0
We calculate terms in parentheses: -(3(2x^2+8)), so:We get rid of parentheses
3(2x^2+8)
We multiply parentheses
6x^2+24
Back to the equation:
-(6x^2+24)
8x^2-6x^2-24+20=0
We add all the numbers together, and all the variables
2x^2-4=0
a = 2; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·2·(-4)
Δ = 32
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*2}=\frac{0-4\sqrt{2}}{4} =-\frac{4\sqrt{2}}{4} =-\sqrt{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*2}=\frac{0+4\sqrt{2}}{4} =\frac{4\sqrt{2}}{4} =\sqrt{2} $
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