Solve 3x*(2x-5)-7*(x+3)=-41

$$3x \times (2x-5)-7 \times (x+3)=-41$$

Solve for x

$$x=2$$

$$x = \frac{5}{3} = 1\frac{2}{3} \approx 1.666666667$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 2x-5.

$$6x^{2}-15x-7\left(x+3\right)=-41$$

Use the distributive property to multiply -7 by x+3.

$$6x^{2}-15x-7x-21=-41$$

Combine -15x and -7x to get -22x.

$$6x^{2}-22x-21=-41$$

Add 41 to both sides.

$$6x^{2}-22x-21+41=0$$

Add -21 and 41 to get 20.

$$6x^{2}-22x+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -22 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\times 6\times 20}}{2\times 6}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\times 6\times 20}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-22\right)±\sqrt{484-24\times 20}}{2\times 6}$$

Multiply -24 times 20.

$$x=\frac{-\left(-22\right)±\sqrt{484-480}}{2\times 6}$$

Add 484 to -480.

$$x=\frac{-\left(-22\right)±\sqrt{4}}{2\times 6}$$

Take the square root of 4.

$$x=\frac{-\left(-22\right)±2}{2\times 6}$$

The opposite of -22 is 22.

$$x=\frac{22±2}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{22±2}{12}$$

Now solve the equation x=\frac{22±2}{12} when ± is plus. Add 22 to 2.

$$x=\frac{24}{12}$$

Divide 24 by 12.

$$x=2$$

Now solve the equation x=\frac{22±2}{12} when ± is minus. Subtract 2 from 22.

$$x=\frac{20}{12}$$

Reduce the fraction \frac{20}{12} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{3}$$

The equation is now solved.

$$x=2 x=\frac{5}{3}$$

Steps for Completing the Square

Use the distributive property to multiply 3x by 2x-5.

$$6x^{2}-15x-7\left(x+3\right)=-41$$

Use the distributive property to multiply -7 by x+3.

$$6x^{2}-15x-7x-21=-41$$

Combine -15x and -7x to get -22x.

$$6x^{2}-22x-21=-41$$

Add 21 to both sides.

$$6x^{2}-22x=-41+21$$

Add -41 and 21 to get -20.

$$6x^{2}-22x=-20$$

Divide both sides by 6.

$$\frac{6x^{2}-22x}{6}=-\frac{20}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{22}{6}\right)x=-\frac{20}{6}$$

Reduce the fraction \frac{-22}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{3}x=-\frac{20}{6}$$

Reduce the fraction \frac{-20}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{11}{3}x=-\frac{10}{3}$$

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-\frac{10}{3}+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-\frac{10}{3}+\frac{121}{36}$$

Add -\frac{10}{3} to \frac{121}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{1}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{1}{6} x-\frac{11}{6}=-\frac{1}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=2 x=\frac{5}{3}$$