What are the solutions to the equation x^2-3x+10=20 ?

The solutions to the equation x^2-3x+10=20 are x=5,x=-2

Find the zeros of x^2-3x+10=20

The zeros of x^2-3x+10=20 are x=5,x=-2

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x^{2} - 3 x + 10 = 20

x = 5, x = - 2

Solution steps

x^{2} - 3 x + 10 = 20

Move

20

to the left side

x^{2} - 3 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 1 \cdot ( - 10 )}{2 \cdot 1}

(- 3)^{2} - 4 \cdot 1 \cdot (- 10) = 7

x_{1, 2} = \frac{- ( - 3 ) \pm 7}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 7}{2 \cdot 1}, x_{2} = \frac{- ( - 3 ) - 7}{2 \cdot 1}

x = \frac{- ( - 3 ) + 7}{2 \cdot 1} : 5

x = \frac{- ( - 3 ) - 7}{2 \cdot 1} : - 2

The solutions to the quadratic equation are:

x = 5, x = - 2

\frac{1}{3} + \frac{1}{6} + \frac{1}{12}

4 5^{\frac{3}{2}} \cdot 4 5^{2}

(\frac{3}{2})^{- 3}

2 j + 2 j = 16

330 \cdot \frac{π}{180}

What are the solutions to the equation x^2-3x+10=20 ?
The solutions to the equation x^2-3x+10=20 are x=5,x=-2
Find the zeros of x^2-3x+10=20
The zeros of x^2-3x+10=20 are x=5,x=-2