Solve -2x^2-3x+1=

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

Factor

$$-2\left(x-\frac{-\sqrt{17}-3}{4}\right)\left(x-\frac{\sqrt{17}-3}{4}\right)$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-2\right)}}{2\left(-2\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-2\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\left(-2\right)}$$

Add 9 to 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{17}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±\sqrt{17}}{-4}$$

Now solve the equation x=\frac{3±\sqrt{17}}{-4} when ± is plus. Add 3 to \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{-4}$$

Divide 3+\sqrt{17} by -4.

$$x=\frac{-\sqrt{17}-3}{4}$$

Now solve the equation x=\frac{3±\sqrt{17}}{-4} when ± is minus. Subtract \sqrt{17} from 3.

$$x=\frac{3-\sqrt{17}}{-4}$$

Divide 3-\sqrt{17} by -4.

$$x=\frac{\sqrt{17}-3}{4}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3-\sqrt{17}}{4} for x_{1} and \frac{-3+\sqrt{17}}{4} for x_{2}.

$$-2x^{2}-3x+1=-2\left(x-\frac{-\sqrt{17}-3}{4}\right)\left(x-\frac{\sqrt{17}-3}{4}\right)$$

Steps Using Direct Factoring Method

Quadratic equations such as this one can be solved by a new direct factoring method that does not require guess work. To use the direct factoring method, the equation must be in the form x^2+Bx+C=0.

$$x ^ 2 +\frac{3}{2}x -\frac{1}{2} = 0$$

Let r and s be the factors for the quadratic equation such that x^2+Bx+C=(x−r)(x−s) where sum of factors (r+s)=−B and the product of factors rs = C

$$r + s = -\frac{3}{2} rs = -\frac{1}{2}$$

Two numbers r and s sum up to -\frac{3}{2} exactly when the average of the two numbers is \frac{1}{2}*-\frac{3}{2} = -\frac{3}{4}. You can also see that the midpoint of r and s corresponds to the axis of symmetry of the parabola represented by the quadratic equation y=x^2+Bx+C. The values of r and s are equidistant from the center by an unknown quantity u. Express r and s with respect to variable u. <div style='padding: 8px'><img src='https://opalmath-gzdabgg4ehffg0hf.b01.azurefd.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = -\frac{3}{4} - u s = -\frac{3}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{1}{2}

$$(-\frac{3}{4} - u) (-\frac{3}{4} + u) = -\frac{1}{2}$$

Simplify by expanding (a -b) (a + b) = a^2 – b^2

$$\frac{9}{16} - u^2 = -\frac{1}{2}$$

Simplify the expression by subtracting \frac{9}{16} on both sides

$$-u^2 = -\frac{1}{2}-\frac{9}{16} = -\frac{17}{16}$$

Simplify the expression by multiplying -1 on both sides and take the square root to obtain the value of unknown variable u

$$u^2 = \frac{17}{16} u = \pm\sqrt{\frac{17}{16}} = \pm \frac{\sqrt{17}}{4}$$

The factors r and s are the solutions to the quadratic equation. Substitute the value of u to compute the r and s.

$$r =-\frac{3}{4} - \frac{\sqrt{17}}{4} = -1.781 s = -\frac{3}{4} + \frac{\sqrt{17}}{4} = 0.281$$

Evaluate

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