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12x^2+38x+27=0
a = 12; b = 38; c = +27;
Δ = b2-4ac
Δ = 382-4·12·27
Δ = 148
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(38)-2\sqrt{37}}{2*12}=\frac{-38-2\sqrt{37}}{24} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2\sqrt{37}}{2*12}=\frac{-38+2\sqrt{37}}{24} $
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