x^2+4x+1.5=2

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Solution for x^2+4x+1.5=2 equation:



x^2+4x+1.5=2
We move all terms to the left:
x^2+4x+1.5-(2)=0
We add all the numbers together, and all the variables
x^2+4x-0.5=0
a = 1; b = 4; c = -0.5;
Δ = b2-4ac
Δ = 42-4·1·(-0.5)
Δ = 18
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:
$\sqrt{\Delta}=\sqrt{18}=\sqrt{9*2}=\sqrt{9}*\sqrt{2}=3\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-3\sqrt{2}}{2*1}=\frac{-4-3\sqrt{2}}{2} $
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+3\sqrt{2}}{2*1}=\frac{-4+3\sqrt{2}}{2} $

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