Solve Delta=[-(4m+1)]^2-4m(3m+3)

$$\Delta = [ - ( 4 m + 1 ) ] ^ { 2 } - 4 m ( 3 m + 3 )$$

Solve for Δ

$$\Delta =\left(2m-1\right)^{2}$$

Solution Steps

To find the opposite of 4m+1, find the opposite of each term.

$$\Delta =\left(-4m-1\right)^{2}-4m\left(3m+3\right)$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4m-1\right)^{2}.

$$\Delta =16m^{2}+8m+1-4m\left(3m+3\right)$$

Use the distributive property to multiply -4m by 3m+3.

$$\Delta =16m^{2}+8m+1-12m^{2}-12m$$

Combine 16m^{2} and -12m^{2} to get 4m^{2}.

$$\Delta =4m^{2}+8m+1-12m$$

Combine 8m and -12m to get -4m.

$$\Delta =4m^{2}-4m+1$$

Solve for m (complex solution)

$$m=\frac{-\sqrt{\Delta }+1}{2}$$

$$m=\frac{\sqrt{\Delta }+1}{2}$$

Solve for m

$$m=\frac{-\sqrt{\Delta }+1}{2}$$

$$m=\frac{\sqrt{\Delta }+1}{2}\text{, }\Delta \geq 0$$