Solve f(x)=-3x^2+5x+7

$$f ( x ) = - 3 x ^ { 2 } + 5 x + 7$$

Factor

$$-3\left(x-\frac{5-\sqrt{109}}{6}\right)\left(x-\frac{\sqrt{109}+5}{6}\right)$$

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.

$$-3x^{2}+5x+7=0$$

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.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-3\right)\times 7}}{2\left(-3\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-3\right)\times 7}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-5±\sqrt{25+12\times 7}}{2\left(-3\right)}$$

Multiply 12 times 7.

$$x=\frac{-5±\sqrt{25+84}}{2\left(-3\right)}$$

Add 25 to 84.

$$x=\frac{-5±\sqrt{109}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-5±\sqrt{109}}{-6}$$

Now solve the equation x=\frac{-5±\sqrt{109}}{-6} when ± is plus. Add -5 to \sqrt{109}.

$$x=\frac{\sqrt{109}-5}{-6}$$

Divide -5+\sqrt{109} by -6.

$$x=\frac{5-\sqrt{109}}{6}$$

Now solve the equation x=\frac{-5±\sqrt{109}}{-6} when ± is minus. Subtract \sqrt{109} from -5.

$$x=\frac{-\sqrt{109}-5}{-6}$$

Divide -5-\sqrt{109} by -6.

$$x=\frac{\sqrt{109}+5}{6}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute \frac{5-\sqrt{109}}{6} for x_{1} and \frac{5+\sqrt{109}}{6} for x_{2}.

$$-3x^{2}+5x+7=-3\left(x-\frac{5-\sqrt{109}}{6}\right)\left(x-\frac{\sqrt{109}+5}{6}\right)$$

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.

$$x ^ 2 -\frac{5}{3}x -\frac{7}{3} = 0$$

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

$$r + s = \frac{5}{3} rs = -\frac{7}{3}$$

Two numbers r and s sum up to \frac{5}{3} exactly when the average of the two numbers is \frac{1}{2}*\frac{5}{3} = \frac{5}{6}. 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.azureedge.net/customsolver/quadraticgraph.png' style='width: 100%;max-width: 700px' /></div>

$$r = \frac{5}{6} - u s = \frac{5}{6} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = -\frac{7}{3}

$$(\frac{5}{6} - u) (\frac{5}{6} + u) = -\frac{7}{3}$$

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

$$\frac{25}{36} - u^2 = -\frac{7}{3}$$

Simplify the expression by subtracting \frac{25}{36} on both sides

$$-u^2 = -\frac{7}{3}-\frac{25}{36} = -\frac{109}{36}$$

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

$$u^2 = \frac{109}{36} u = \pm\sqrt{\frac{109}{36}} = \pm \frac{\sqrt{109}}{6}$$

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

$$r =\frac{5}{6} - \frac{\sqrt{109}}{6} = -0.907 s = \frac{5}{6} + \frac{\sqrt{109}}{6} = 2.573$$

Evaluate

$$7+5x-3x^{2}$$