Solve x^2-5x+3y=20 | Microsoft Math Solver

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Solve for x

x = \frac{1 0 5 - 1 2 y + 5}{2}

x = \frac{- 1 0 5 - 1 2 y + 5}{2}, y \leq \frac{3 5}{4}

Steps Using the Quadratic Formula

Steps for Completing the Square

Steps Using the Quadratic Formula

x^{2} - 5 x + 3 y = 20

All equations of the form

a x^{2} + b x + c = 0

can be solved using the quadratic formula:

\frac{- b \pm b ^{2} - 4 a c}{2 a}

. The quadratic formula gives two solutions, one when

\pm

is addition and one when it is subtraction.

x^{2} - 5 x + 3 y = 20

Subtract

20

from both sides of the equation.

x^{2} - 5 x + 3 y - 20 = 20 - 20

Subtracting

20

from itself leaves

0

.

x^{2} - 5 x + 3 y - 20 = 0

This equation is in standard form:

a x^{2} + b x + c = 0

. Substitute

1

for

a

,

- 5

for

b

, and

3 y - 20

for

c

in the quadratic formula,

\frac{- b \pm b ^{2} - 4 a c}{2 a}

.

x = \frac{- ( - 5 ) \pm ( - 5 ) ^{2} - 4 ( 3 y - 2 0 )}{2}

Square

- 5

.

x = \frac{- ( - 5 ) \pm 2 5 - 4 ( 3 y - 2 0 )}{2}

Multiply

- 4

times

3 y - 20

.

x = \frac{- ( - 5 ) \pm 2 5 + 8 0 - 1 2 y}{2}

Add

25

to

- 12 y + 80

.

x = \frac{- ( - 5 ) \pm 1 0 5 - 1 2 y}{2}

The opposite of

- 5

is

5

.

x = \frac{5 \pm 1 0 5 - 1 2 y}{2}

Now solve the equation

x = \frac{5 \pm 1 0 5 - 1 2 y}{2}

when

\pm

is plus. Add

5

to

105 - 12 y

.

x = \frac{1 0 5 - 1 2 y + 5}{2}

Now solve the equation

x = \frac{5 \pm 1 0 5 - 1 2 y}{2}

when

\pm

is minus. Subtract

105 - 12 y

from

5

.

x = \frac{- 1 0 5 - 1 2 y + 5}{2}

The equation is now solved.

x = \frac{1 0 5 - 1 2 y + 5}{2}

x = \frac{- 1 0 5 - 1 2 y + 5}{2}

Solve for y

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

Steps for Solving Linear Equation

x^{2} - 5 x + 3 y = 20

Subtract

x^{2}

from both sides.

- 5 x + 3 y = 20 - x^{2}

Add

5 x

to both sides.

3 y = 20 - x^{2} + 5 x

The equation is in standard form.

3 y = 20 + 5 x - x^{2}

Divide both sides by

3

.

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

Dividing by

3

undoes the multiplication by

3

.

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

Solve for x (complex solution)

x = \frac{1 0 5 - 1 2 y + 5}{2}

x = \frac{- 1 0 5 - 1 2 y + 5}{2}

Graph

Quiz

10 problems similar to:

x^{2} - 5 x + 3 y = 20

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What are the points of intersection of

y = - 2 x^{2} - 5 x + 3

y = - 2 x + 3

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(0, 3), and, (- \frac{3}{2}, 6)

. Explanation: To find the pts. of intersection of these two curves, we have to solve their eqns. ...

How do you find the vertex and axis of symmetry of

y = - 4 x^{2} + 8 x + 13

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Vertex (1, 17) Explanation: x-coordinate of vertex, and of the axis of symmetry:

x = - \frac{b}{2 a} = - \frac{8}{- 8} = 1

y-coordinate of vertex: y(1) = -4 + 8 + 13 = ...

How do you write the quadratic in vertex form given

y = - 4 x^{2} - 5 x + 3

https://socratic.org/questions/how-do-you-write-the-quadratic-in-vertex-form-given-y-4x-2-5x-3

y = (- 4) (x + \frac{5}{8})^{2} + \frac{7 3}{1 6}

Explanation: The general vertex form of a quadratic is

y = m (x - a)^{2} + b

How do you find the zeros, real and imaginary, of

y = - 7 x^{2} - 5 x + 3

using the quadratic formula?

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The zeroes (or roots, as they are also called) of

y = - 7 x^{2} - 5 x + 3

x = \frac{5 \pm 1 0 9}{- 1 4}

Explanation: The problem says to use ...

How do you find the vertex and the intercepts for

y = x^{2} + 10 x + 22

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vertex is (-5,-3). It intersects with x-axis in two points (-5, -+sqrt3) & y-intercept 22. Explanation: We have, by completing square,

y = x^{2} + 10 x + 25 - 3

y + 3 = (x + 5)^{2}

How do you find the vertex and the intercepts for

y = - x^{2} + 8 x + 16

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(4, 32)

(0, 16)

Explanation: Given -

y = - x^{2} + 8 x + 16

x = \frac{- b}{2 a} = \frac{- 8}{2 \times ( - 1 )} = \frac{- 8}{- 2} = 4

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x^{2}-5x+3y=20

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^{2}-5x+3y-20=20-20

Subtract 20 from both sides of the equation.

x^{2}-5x+3y-20=0

Subtracting 20 from itself leaves 0.

x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(3y-20\right)}}{2}

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and 3y-20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

x=\frac{-\left(-5\right)±\sqrt{25-4\left(3y-20\right)}}{2}

Square -5.

x=\frac{-\left(-5\right)±\sqrt{25+80-12y}}{2}

Multiply -4 times 3y-20.

x=\frac{-\left(-5\right)±\sqrt{105-12y}}{2}

Add 25 to -12y+80.

x=\frac{5±\sqrt{105-12y}}{2}

The opposite of -5 is 5.

x=\frac{\sqrt{105-12y}+5}{2}

Now solve the equation x=\frac{5±\sqrt{105-12y}}{2} when ± is plus. Add 5 to \sqrt{105-12y}.

x=\frac{-\sqrt{105-12y}+5}{2}

Now solve the equation x=\frac{5±\sqrt{105-12y}}{2} when ± is minus. Subtract \sqrt{105-12y} from 5.

x=\frac{\sqrt{105-12y}+5}{2} x=\frac{-\sqrt{105-12y}+5}{2}

The equation is now solved.

x^{2}-5x+3y=20

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.

x^{2}-5x+3y-3y=20-3y

Subtract 3y from both sides of the equation.

x^{2}-5x=20-3y

Subtracting 3y from itself leaves 0.

x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=20-3y+\left(-\frac{5}{2}\right)^{2}

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}=-2.5. Then add the square of -\frac{5}{2}=-2.5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

x^{2}-5x+\frac{25}{4}=20-3y+\frac{25}{4}

Square -\frac{5}{2}=-2.5 by squaring both the numerator and the denominator of the fraction.

x^{2}-5x+\frac{25}{4}=\frac{105}{4}-3y

Add 20-3y to \frac{25}{4}=6.25.

\left(x-\frac{5}{2}\right)^{2}=\frac{105}{4}-3y

Factor x^{2}-5x+\frac{25}{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{5}{2}\right)^{2}}=\sqrt{\frac{105}{4}-3y}

Take the square root of both sides of the equation.

x-\frac{5}{2}=\frac{\sqrt{105-12y}}{2} x-\frac{5}{2}=-\frac{\sqrt{105-12y}}{2}

Simplify.

x=\frac{\sqrt{105-12y}+5}{2} x=\frac{-\sqrt{105-12y}+5}{2}

Add \frac{5}{2}=2.5 to both sides of the equation.

-5x+3y=20-x^{2}

Subtract x^{2} from both sides.

3y=20-x^{2}+5x

Add 5x to both sides.

3y=20+5x-x^{2}

The equation is in standard form.

\frac{3y}{3}=\frac{20+5x-x^{2}}{3}

Divide both sides by 3.

y=\frac{20+5x-x^{2}}{3}

Dividing by 3 undoes the multiplication by 3.

Similar Problems

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x^{2} - 5 x + 3 y = 20

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

2 x^{2} + 12 x + 40 = 0

\frac{1}{3} = m + \frac{m - 1}{m}

\frac{2}{b - 3} - \frac{6}{2 b + 1} = 4

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