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3x^2+2x+16=7x^2+2x
We move all terms to the left:
3x^2+2x+16-(7x^2+2x)=0
We get rid of parentheses
3x^2-7x^2+2x-2x+16=0
We add all the numbers together, and all the variables
-4x^2+16=0
a = -4; b = 0; c = +16;
Δ = b2-4ac
Δ = 02-4·(-4)·16
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*-4}=\frac{-16}{-8} =+2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*-4}=\frac{16}{-8} =-2 $
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