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4x+11x^2=195
We move all terms to the left:
4x+11x^2-(195)=0
a = 11; b = 4; c = -195;
Δ = b2-4ac
Δ = 42-4·11·(-195)
Δ = 8596
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{8596}=\sqrt{4*2149}=\sqrt{4}*\sqrt{2149}=2\sqrt{2149}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-2\sqrt{2149}}{2*11}=\frac{-4-2\sqrt{2149}}{22} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+2\sqrt{2149}}{2*11}=\frac{-4+2\sqrt{2149}}{22} $
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