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(8x+x)(x)=62
We move all terms to the left:
(8x+x)(x)-(62)=0
We add all the numbers together, and all the variables
(+9x)x-62=0
We multiply parentheses
9x^2-62=0
a = 9; b = 0; c = -62;
Δ = b2-4ac
Δ = 02-4·9·(-62)
Δ = 2232
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2232}=\sqrt{36*62}=\sqrt{36}*\sqrt{62}=6\sqrt{62}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{62}}{2*9}=\frac{0-6\sqrt{62}}{18} =-\frac{6\sqrt{62}}{18} =-\frac{\sqrt{62}}{3} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{62}}{2*9}=\frac{0+6\sqrt{62}}{18} =\frac{6\sqrt{62}}{18} =\frac{\sqrt{62}}{3} $
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