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3x^2-9x+20.25=26.25
We move all terms to the left:
3x^2-9x+20.25-(26.25)=0
We add all the numbers together, and all the variables
3x^2-9x-6=0
a = 3; b = -9; c = -6;
Δ = b2-4ac
Δ = -92-4·3·(-6)
Δ = 153
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{17}}{2*3}=\frac{9-3\sqrt{17}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{17}}{2*3}=\frac{9+3\sqrt{17}}{6} $
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