Solve 2^2x^2+3x-2

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

Factor

$$4\left(x-\frac{-\sqrt{41}-3}{8}\right)\left(x-\frac{\sqrt{41}-3}{8}\right)$$

Steps Using the Quadratic Formula

Calculate 2 to the power of 2 and get 4.

$$factor(4x^{2}+3x-2)$$

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.

$$4x^{2}+3x-2=0$$

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.

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

Square 3.

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

Multiply -4 times 4.

$$x=\frac{-3±\sqrt{9-16\left(-2\right)}}{2\times 4}$$

Multiply -16 times -2.

$$x=\frac{-3±\sqrt{9+32}}{2\times 4}$$

Add 9 to 32.

$$x=\frac{-3±\sqrt{41}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-3±\sqrt{41}}{8}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{8} when ± is plus. Add -3 to \sqrt{41}.

$$x=\frac{\sqrt{41}-3}{8}$$

Now solve the equation x=\frac{-3±\sqrt{41}}{8} when ± is minus. Subtract \sqrt{41} from -3.

$$x=\frac{-\sqrt{41}-3}{8}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{41}}{8} for x_{1} and \frac{-3-\sqrt{41}}{8} for x_{2}.

$$4x^{2}+3x-2=4\left(x-\frac{\sqrt{41}-3}{8}\right)\left(x-\frac{-\sqrt{41}-3}{8}\right)$$

Evaluate

$$4x^{2}+3x-2$$

Solution Steps

Calculate 2 to the power of 2 and get 4.

$$4x^{2}+3x-2$$