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4w^2-12+9=0
We add all the numbers together, and all the variables
4w^2-3=0
a = 4; b = 0; c = -3;
Δ = b2-4ac
Δ = 02-4·4·(-3)
Δ = 48
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$w_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*4}=\frac{0-4\sqrt{3}}{8} =-\frac{4\sqrt{3}}{8} =-\frac{\sqrt{3}}{2} $$w_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*4}=\frac{0+4\sqrt{3}}{8} =\frac{4\sqrt{3}}{8} =\frac{\sqrt{3}}{2} $
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