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03x^2+36x+81=03
We move all terms to the left:
03x^2+36x+81-(03)=0
We add all the numbers together, and all the variables
03x^2+36x+78=0
a = 03; b = 36; c = +78;
Δ = b2-4ac
Δ = 362-4·03·78
Δ = 360
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{360}=\sqrt{36*10}=\sqrt{36}*\sqrt{10}=6\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-6\sqrt{10}}{2*03}=\frac{-36-6\sqrt{10}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+6\sqrt{10}}{2*03}=\frac{-36+6\sqrt{10}}{6} $
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