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6x^2+21x-45=3
We move all terms to the left:
6x^2+21x-45-(3)=0
We add all the numbers together, and all the variables
6x^2+21x-48=0
a = 6; b = 21; c = -48;
Δ = b2-4ac
Δ = 212-4·6·(-48)
Δ = 1593
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{1593}=\sqrt{9*177}=\sqrt{9}*\sqrt{177}=3\sqrt{177}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{177}}{2*6}=\frac{-21-3\sqrt{177}}{12} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{177}}{2*6}=\frac{-21+3\sqrt{177}}{12} $
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