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4x^2-65=-16x
We move all terms to the left:
4x^2-65-(-16x)=0
We get rid of parentheses
4x^2+16x-65=0
a = 4; b = 16; c = -65;
Δ = b2-4ac
Δ = 162-4·4·(-65)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-36}{2*4}=\frac{-52}{8} =-6+1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+36}{2*4}=\frac{20}{8} =2+1/2 $
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