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3x^2+12x-62=7
We move all terms to the left:
3x^2+12x-62-(7)=0
We add all the numbers together, and all the variables
3x^2+12x-69=0
a = 3; b = 12; c = -69;
Δ = b2-4ac
Δ = 122-4·3·(-69)
Δ = 972
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{972}=\sqrt{324*3}=\sqrt{324}*\sqrt{3}=18\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-18\sqrt{3}}{2*3}=\frac{-12-18\sqrt{3}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+18\sqrt{3}}{2*3}=\frac{-12+18\sqrt{3}}{6} $
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