Solve 10x=25-3x^2

$$10 x = 25 - 3 x ^ { 2 }$$

Solve for x

$$x=-5$$

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

Steps Using Factoring By Grouping

Subtract 25 from both sides.

$$10x-25=-3x^{2}$$

Add 3x^{2} to both sides.

$$10x-25+3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$3x^{2}+10x-25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-25. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\left(-25\right)=-75$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -75.

$$-1,75 -3,25 -5,15$$

Calculate the sum for each pair.

$$-1+75=74 -3+25=22 -5+15=10$$

The solution is the pair that gives sum 10.

$$a=-5 b=15$$

Rewrite 3x^{2}+10x-25 as \left(3x^{2}-5x\right)+\left(15x-25\right).

$$\left(3x^{2}-5x\right)+\left(15x-25\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(3x-5\right)+5\left(3x-5\right)$$

Factor out common term 3x-5 by using distributive property.

$$\left(3x-5\right)\left(x+5\right)$$

To find equation solutions, solve 3x-5=0 and x+5=0.

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

Steps Using the Quadratic Formula

Subtract 25 from both sides.

$$10x-25=-3x^{2}$$

Add 3x^{2} to both sides.

$$10x-25+3x^{2}=0$$

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.

$$3x^{2}+10x-25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 10 for b, and -25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-10±\sqrt{10^{2}-4\times 3\left(-25\right)}}{2\times 3}$$

Square 10.

$$x=\frac{-10±\sqrt{100-4\times 3\left(-25\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-10±\sqrt{100-12\left(-25\right)}}{2\times 3}$$

Multiply -12 times -25.

$$x=\frac{-10±\sqrt{100+300}}{2\times 3}$$

Add 100 to 300.

$$x=\frac{-10±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$x=\frac{-10±20}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-10±20}{6}$$

Now solve the equation x=\frac{-10±20}{6} when ± is plus. Add -10 to 20.

$$x=\frac{10}{6}$$

Reduce the fraction \frac{10}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-10±20}{6} when ± is minus. Subtract 20 from -10.

$$x=-\frac{30}{6}$$

Divide -30 by 6.

$$x=-5$$

The equation is now solved.

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

Steps for Completing the Square

Add 3x^{2} to both sides.

$$10x+3x^{2}=25$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$3x^{2}+10x=25$$

Divide both sides by 3.

$$\frac{3x^{2}+10x}{3}=\frac{25}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{10}{3}x=\frac{25}{3}$$

Divide \frac{10}{3}, the coefficient of the x term, by 2 to get \frac{5}{3}. Then add the square of \frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{25}{3}+\left(\frac{5}{3}\right)^{2}$$

Square \frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{25}{3}+\frac{25}{9}$$

Add \frac{25}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{100}{9}$$

Factor x^{2}+\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{10}{3} x+\frac{5}{3}=-\frac{10}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

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