Solve 2x^2+3x+(1-a)=0

$$2 x ^ { 2 } + 3 x + ( 1 - a ) = 0$$

Solve for a

$$a=\left(x+1\right)\left(2x+1\right)$$

Steps for Solving Linear Equation

Subtract 2x^{2} from both sides. Anything subtracted from zero gives its negation.

$$3x+1-a=-2x^{2}$$

Subtract 3x from both sides.

$$1-a=-2x^{2}-3x$$

Subtract 1 from both sides.

$$-a=-2x^{2}-3x-1$$

Divide both sides by -1.

$$\frac{-a}{-1}=-\frac{\left(x+1\right)\left(2x+1\right)}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$a=-\frac{\left(x+1\right)\left(2x+1\right)}{-1}$$

Divide -\left(1+x\right)\left(1+2x\right) by -1.

$$a=\left(x+1\right)\left(2x+1\right)$$

Solve for x (complex solution)

$$x=\frac{\sqrt{8a+1}-3}{4}$$

$$x=\frac{-\sqrt{8a+1}-3}{4}$$

Solve for x

$$x=\frac{\sqrt{8a+1}-3}{4}$$

$$x=\frac{-\sqrt{8a+1}-3}{4}\text{, }a\geq -\frac{1}{8}$$