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4x^2+16x=209
We move all terms to the left:
4x^2+16x-(209)=0
a = 4; b = 16; c = -209;
Δ = b2-4ac
Δ = 162-4·4·(-209)
Δ = 3600
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{3600}=60$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-60}{2*4}=\frac{-76}{8} =-9+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+60}{2*4}=\frac{44}{8} =5+1/2 $
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