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4x^2+21x-216=0
a = 4; b = 21; c = -216;
Δ = b2-4ac
Δ = 212-4·4·(-216)
Δ = 3897
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{3897}=\sqrt{9*433}=\sqrt{9}*\sqrt{433}=3\sqrt{433}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(21)-3\sqrt{433}}{2*4}=\frac{-21-3\sqrt{433}}{8} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(21)+3\sqrt{433}}{2*4}=\frac{-21+3\sqrt{433}}{8} $
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