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3x^2+2x^2=57
We move all terms to the left:
3x^2+2x^2-(57)=0
We add all the numbers together, and all the variables
5x^2-57=0
a = 5; b = 0; c = -57;
Δ = b2-4ac
Δ = 02-4·5·(-57)
Δ = 1140
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1140}=\sqrt{4*285}=\sqrt{4}*\sqrt{285}=2\sqrt{285}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{285}}{2*5}=\frac{0-2\sqrt{285}}{10} =-\frac{2\sqrt{285}}{10} =-\frac{\sqrt{285}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{285}}{2*5}=\frac{0+2\sqrt{285}}{10} =\frac{2\sqrt{285}}{10} =\frac{\sqrt{285}}{5} $
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