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2x^2+12x+36=35
We move all terms to the left:
2x^2+12x+36-(35)=0
We add all the numbers together, and all the variables
2x^2+12x+1=0
a = 2; b = 12; c = +1;
Δ = b2-4ac
Δ = 122-4·2·1
Δ = 136
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-2\sqrt{34}}{2*2}=\frac{-12-2\sqrt{34}}{4} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+2\sqrt{34}}{2*2}=\frac{-12+2\sqrt{34}}{4} $
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