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x^{2}-16=2x^{2}-3\left(x+\frac{20}{3}\right)
Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.
x^{2}-16=2x^{2}-3x-20
Use the distributive property to multiply -3 by x+\frac{20}{3}.
x^{2}-16-2x^{2}=-3x-20
Subtract 2x^{2} from both sides.
-x^{2}-16=-3x-20
Combine x^{2} and -2x^{2} to get -x^{2}.
-x^{2}-16+3x=-20
Add 3x to both sides.
-x^{2}-16+3x+20=0
Add 20 to both sides.
-x^{2}+4+3x=0
Add -16 and 20 to get 4.
-x^{2}+3x+4=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=3 ab=-4=-4
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+4. To find a and b, set up a system to be solved.
-1,4 -2,2
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 -4.
-1+4=3 -2+2=0
Calculate the sum for each pair.
a=4 b=-1
The solution is the pair that gives sum 3.
\left(-x^{2}+4x\right)+\left(-x+4\right)
Rewrite -x^{2}+3x+4 as \left(-x^{2}+4x\right)+\left(-x+4\right).
-x\left(x-4\right)-\left(x-4\right)
Factor out -x in the first and -1 in the second group.
\left(x-4\right)\left(-x-1\right)
Factor out common term x-4 by using distributive property.
x=4 x=-1
To find equation solutions, solve x-4=0 and -x-1=0.
x^{2}-16=2x^{2}-3\left(x+\frac{20}{3}\right)
Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.
x^{2}-16=2x^{2}-3x-20
Use the distributive property to multiply -3 by x+\frac{20}{3}.
x^{2}-16-2x^{2}=-3x-20
Subtract 2x^{2} from both sides.
-x^{2}-16=-3x-20
Combine x^{2} and -2x^{2} to get -x^{2}.
-x^{2}-16+3x=-20
Add 3x to both sides.
-x^{2}-16+3x+20=0
Add 20 to both sides.
-x^{2}+4+3x=0
Add -16 and 20 to get 4.
-x^{2}+3x+4=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\left(-1\right)\times 4}}{2\left(-1\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 3 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-3±\sqrt{9-4\left(-1\right)\times 4}}{2\left(-1\right)}
Square 3.
x=\frac{-3±\sqrt{9+4\times 4}}{2\left(-1\right)}
Multiply -4 times -1.
x=\frac{-3±\sqrt{9+16}}{2\left(-1\right)}
Multiply 4 times 4.
x=\frac{-3±\sqrt{25}}{2\left(-1\right)}
Add 9 to 16.
x=\frac{-3±5}{2\left(-1\right)}
Take the square root of 25.
x=\frac{-3±5}{-2}
Multiply 2 times -1.
x=\frac{2}{-2}
Now solve the equation x=\frac{-3±5}{-2} when ± is plus. Add -3 to 5.
x=-1
Divide 2 by -2.
x=-\frac{8}{-2}
Now solve the equation x=\frac{-3±5}{-2} when ± is minus. Subtract 5 from -3.
x=4
Divide -8 by -2.
x=-1 x=4
The equation is now solved.
x^{2}-16=2x^{2}-3\left(x+\frac{20}{3}\right)
Consider \left(x-4\right)\left(x+4\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 4.
x^{2}-16=2x^{2}-3x-20
Use the distributive property to multiply -3 by x+\frac{20}{3}.
x^{2}-16-2x^{2}=-3x-20
Subtract 2x^{2} from both sides.
-x^{2}-16=-3x-20
Combine x^{2} and -2x^{2} to get -x^{2}.
-x^{2}-16+3x=-20
Add 3x to both sides.
-x^{2}+3x=-20+16
Add 16 to both sides.
-x^{2}+3x=-4
Add -20 and 16 to get -4.
\frac{-x^{2}+3x}{-1}=-\frac{4}{-1}
Divide both sides by -1.
x^{2}+\frac{3}{-1}x=-\frac{4}{-1}
Dividing by -1 undoes the multiplication by -1.
x^{2}-3x=-\frac{4}{-1}
Divide 3 by -1.
x^{2}-3x=4
Divide -4 by -1.
x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=4+\left(-\frac{3}{2}\right)^{2}
Divide -3, the coefficient of the x term, by 2 to get -\frac{3}{2}. Then add the square of -\frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-3x+\frac{9}{4}=4+\frac{9}{4}
Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}-3x+\frac{9}{4}=\frac{25}{4}
Add 4 to \frac{9}{4}.
\left(x-\frac{3}{2}\right)^{2}=\frac{25}{4}
Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.
\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{25}{4}}
Take the square root of both sides of the equation.
x-\frac{3}{2}=\frac{5}{2} x-\frac{3}{2}=-\frac{5}{2}
Simplify.
x=4 x=-1
Add \frac{3}{2} to both sides of the equation.