Solve (5x^2-4x+3)+(2x^2+7x-9)

$$( 5 x ^ { 2 } - 4 x + 3 ) + ( 2 x ^ { 2 } + 7 x - 9 )$$

Evaluate

$$7x^{2}+3x-6$$

Solution Steps

Combine 5x^{2} and 2x^{2} to get 7x^{2}.

$$7x^{2}-4x+3+7x-9$$

Combine -4x and 7x to get 3x.

$$7x^{2}+3x+3-9$$

Subtract 9 from 3 to get -6.

$$7x^{2}+3x-6$$

Factor

$$7\left(x-\frac{-\sqrt{177}-3}{14}\right)\left(x-\frac{\sqrt{177}-3}{14}\right)$$

Steps Using the Quadratic Formula

Combine 5x^{2} and 2x^{2} to get 7x^{2}.

$$factor(7x^{2}-4x+3+7x-9)$$

Combine -4x and 7x to get 3x.

$$factor(7x^{2}+3x+3-9)$$

Subtract 9 from 3 to get -6.

$$factor(7x^{2}+3x-6)$$

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.

$$7x^{2}+3x-6=0$$

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.

$$x=\frac{-3±\sqrt{3^{2}-4\times 7\left(-6\right)}}{2\times 7}$$

Square 3.

$$x=\frac{-3±\sqrt{9-4\times 7\left(-6\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-3±\sqrt{9-28\left(-6\right)}}{2\times 7}$$

Multiply -28 times -6.

$$x=\frac{-3±\sqrt{9+168}}{2\times 7}$$

Add 9 to 168.

$$x=\frac{-3±\sqrt{177}}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{-3±\sqrt{177}}{14}$$

Now solve the equation x=\frac{-3±\sqrt{177}}{14} when ± is plus. Add -3 to \sqrt{177}.

$$x=\frac{\sqrt{177}-3}{14}$$

Now solve the equation x=\frac{-3±\sqrt{177}}{14} when ± is minus. Subtract \sqrt{177} from -3.

$$x=\frac{-\sqrt{177}-3}{14}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{-3+\sqrt{177}}{14} for x_{1} and \frac{-3-\sqrt{177}}{14} for x_{2}.

$$7x^{2}+3x-6=7\left(x-\frac{\sqrt{177}-3}{14}\right)\left(x-\frac{-\sqrt{177}-3}{14}\right)$$