Solve D=(2x+3)^2+(x-5)(2x+3)

$$D = ( 2 x + 3 ) ^ { 2 } + ( x - 5 ) ( 2 x + 3 )$$

Solve for D

$$D=\left(3x-2\right)\left(2x+3\right)$$

Solution Steps

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+3\right)^{2}.

$$D=4x^{2}+12x+9+\left(x-5\right)\left(2x+3\right)$$

Use the distributive property to multiply x-5 by 2x+3 and combine like terms.

$$D=4x^{2}+12x+9+2x^{2}-7x-15$$

Combine 4x^{2} and 2x^{2} to get 6x^{2}.

$$D=6x^{2}+12x+9-7x-15$$

Combine 12x and -7x to get 5x.

$$D=6x^{2}+5x+9-15$$

Subtract 15 from 9 to get -6.

$$D=6x^{2}+5x-6$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{24D+169}-5}{12}$$

$$x=\frac{\sqrt{24D+169}-5}{12}$$

Solve for x

$$x=\frac{-\sqrt{24D+169}-5}{12}$$

$$x=\frac{\sqrt{24D+169}-5}{12}\text{, }D\geq -\frac{169}{24}$$