Solve -4x^2+10x-6=

$$- 4 x ^ { 2 } + 10 x - 6 =$$

Factor

$$2\left(1-x\right)\left(2x-3\right)$$

Solution Steps

Factor out 2.

$$2\left(-2x^{2}+5x-3\right)$$

Consider -2x^{2}+5x-3. Factor the expression by grouping. First, the expression needs to be rewritten as -2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=5 ab=-2\left(-3\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -2x^{2}+5x-3 as \left(-2x^{2}+3x\right)+\left(2x-3\right).

$$\left(-2x^{2}+3x\right)+\left(2x-3\right)$$

Factor out -x in -2x^{2}+3x.

$$-x\left(2x-3\right)+2x-3$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(-x+1\right)$$

Rewrite the complete factored expression.

$$2\left(2x-3\right)\left(-x+1\right)$$

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.

$$-4x^{2}+10x-6=0$$

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.

$$x=\frac{-10±\sqrt{10^{2}-4\left(-4\right)\left(-6\right)}}{2\left(-4\right)}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\left(-4\right)\left(-6\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-10±\sqrt{100+16\left(-6\right)}}{2\left(-4\right)}$$

Multiply 16 times -6.

$$x=\frac{-10±\sqrt{100-96}}{2\left(-4\right)}$$

Add 100 to -96.

$$x=\frac{-10±\sqrt{4}}{2\left(-4\right)}$$

Take the square root of 4.

$$x=\frac{-10±2}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-10±2}{-8}$$

Now solve the equation x=\frac{-10±2}{-8} when ± is plus. Add -10 to 2.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

Now solve the equation x=\frac{-10±2}{-8} when ± is minus. Subtract 2 from -10.

$$x=-\frac{12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{3}{2}$$

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

$$-4x^{2}+10x-6=-4\left(x-1\right)\left(x-\frac{3}{2}\right)$$

Subtract \frac{3}{2} from x by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$-4x^{2}+10x-6=-4\left(x-1\right)\times \frac{-2x+3}{-2}$$

Cancel out 2, the greatest common factor in -4 and 2.

$$-4x^{2}+10x-6=2\left(x-1\right)\left(-2x+3\right)$$

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.

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

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

$$r + s = \frac{5}{2} rs = \frac{3}{2}$$

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

$$r = \frac{5}{4} - u s = \frac{5}{4} + u$$

To solve for unknown quantity u, substitute these in the product equation rs = \frac{3}{2}

$$(\frac{5}{4} - u) (\frac{5}{4} + u) = \frac{3}{2}$$

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

$$\frac{25}{16} - u^2 = \frac{3}{2}$$

Simplify the expression by subtracting \frac{25}{16} on both sides

$$-u^2 = \frac{3}{2}-\frac{25}{16} = -\frac{1}{16}$$

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

$$u^2 = \frac{1}{16} u = \pm\sqrt{\frac{1}{16}} = \pm \frac{1}{4}$$

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

$$r =\frac{5}{4} - \frac{1}{4} = 1 s = \frac{5}{4} + \frac{1}{4} = 1.500$$

Evaluate

$$-4x^{2}+10x-6$$