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9x^2+16=21x+28
We move all terms to the left:
9x^2+16-(21x+28)=0
We get rid of parentheses
9x^2-21x-28+16=0
We add all the numbers together, and all the variables
9x^2-21x-12=0
a = 9; b = -21; c = -12;
Δ = b2-4ac
Δ = -212-4·9·(-12)
Δ = 873
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{873}=\sqrt{9*97}=\sqrt{9}*\sqrt{97}=3\sqrt{97}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-21)-3\sqrt{97}}{2*9}=\frac{21-3\sqrt{97}}{18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-21)+3\sqrt{97}}{2*9}=\frac{21+3\sqrt{97}}{18} $
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