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4x^2+25=3x^2+169
We move all terms to the left:
4x^2+25-(3x^2+169)=0
We get rid of parentheses
4x^2-3x^2-169+25=0
We add all the numbers together, and all the variables
x^2-144=0
a = 1; b = 0; c = -144;
Δ = b2-4ac
Δ = 02-4·1·(-144)
Δ = 576
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{576}=24$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-24}{2*1}=\frac{-24}{2} =-12 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+24}{2*1}=\frac{24}{2} =12 $
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