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3x^2+7x^2=34^2
We move all terms to the left:
3x^2+7x^2-(34^2)=0
We add all the numbers together, and all the variables
10x^2-1156=0
a = 10; b = 0; c = -1156;
Δ = b2-4ac
Δ = 02-4·10·(-1156)
Δ = 46240
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{46240}=\sqrt{4624*10}=\sqrt{4624}*\sqrt{10}=68\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-68\sqrt{10}}{2*10}=\frac{0-68\sqrt{10}}{20} =-\frac{68\sqrt{10}}{20} =-\frac{17\sqrt{10}}{5} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+68\sqrt{10}}{2*10}=\frac{0+68\sqrt{10}}{20} =\frac{68\sqrt{10}}{20} =\frac{17\sqrt{10}}{5} $
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