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9x^2-3x+7=25
We move all terms to the left:
9x^2-3x+7-(25)=0
We add all the numbers together, and all the variables
9x^2-3x-18=0
a = 9; b = -3; c = -18;
Δ = b2-4ac
Δ = -32-4·9·(-18)
Δ = 657
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{657}=\sqrt{9*73}=\sqrt{9}*\sqrt{73}=3\sqrt{73}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3\sqrt{73}}{2*9}=\frac{3-3\sqrt{73}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3\sqrt{73}}{2*9}=\frac{3+3\sqrt{73}}{18} $
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