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4a(a-3)=22
We move all terms to the left:
4a(a-3)-(22)=0
We multiply parentheses
4a^2-12a-22=0
a = 4; b = -12; c = -22;
Δ = b2-4ac
Δ = -122-4·4·(-22)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{31}}{2*4}=\frac{12-4\sqrt{31}}{8} $$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{31}}{2*4}=\frac{12+4\sqrt{31}}{8} $
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