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4x^2+3x^2=65
We move all terms to the left:
4x^2+3x^2-(65)=0
We add all the numbers together, and all the variables
7x^2-65=0
a = 7; b = 0; c = -65;
Δ = b2-4ac
Δ = 02-4·7·(-65)
Δ = 1820
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1820}=\sqrt{4*455}=\sqrt{4}*\sqrt{455}=2\sqrt{455}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{455}}{2*7}=\frac{0-2\sqrt{455}}{14} =-\frac{2\sqrt{455}}{14} =-\frac{\sqrt{455}}{7} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{455}}{2*7}=\frac{0+2\sqrt{455}}{14} =\frac{2\sqrt{455}}{14} =\frac{\sqrt{455}}{7} $
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