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2x(2x+1)=2x+10
We move all terms to the left:
2x(2x+1)-(2x+10)=0
We multiply parentheses
4x^2+2x-(2x+10)=0
We get rid of parentheses
4x^2+2x-2x-10=0
We add all the numbers together, and all the variables
4x^2-10=0
a = 4; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·4·(-10)
Δ = 160
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{160}=\sqrt{16*10}=\sqrt{16}*\sqrt{10}=4\sqrt{10}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{10}}{2*4}=\frac{0-4\sqrt{10}}{8} =-\frac{4\sqrt{10}}{8} =-\frac{\sqrt{10}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{10}}{2*4}=\frac{0+4\sqrt{10}}{8} =\frac{4\sqrt{10}}{8} =\frac{\sqrt{10}}{2} $
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