What are the solutions to the equation x^2-3(2x-5)=10 ?

The solutions to the equation x^2-3(2x-5)=10 are x=5,x=1

Find the zeros of x^2-3(2x-5)=10

The zeros of x^2-3(2x-5)=10 are x=5,x=1

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x^{2} - 3 (2 x - 5) = 10

x = 5, x = 1

Solution steps

x^{2} - 3 (2 x - 5) = 10

Expand

x^{2} - 3 (2 x - 5) : x^{2} - 6 x + 15

x^{2} - 6 x + 15 = 10

Move

10

to the left side

x^{2} - 6 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 6 ) \pm ( - 6 ) ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

(- 6)^{2} - 4 \cdot 1 \cdot 5 = 4

x_{1, 2} = \frac{- ( - 6 ) \pm 4}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- ( - 6 ) + 4}{2 \cdot 1}, x_{2} = \frac{- ( - 6 ) - 4}{2 \cdot 1}

x = \frac{- ( - 6 ) + 4}{2 \cdot 1} : 5

x = \frac{- ( - 6 ) - 4}{2 \cdot 1} : 1

The solutions to the quadratic equation are:

x = 5, x = 1

x (2 - x) = - 2

x, x^{2} - k x + 1681 = 0

(2 x - 1)^{2} = (x + 1)^{2}

(x - 9)^{2} = - 36

5 - 2 x^{2} = 6 x

What are the solutions to the equation x^2-3(2x-5)=10 ?
The solutions to the equation x^2-3(2x-5)=10 are x=5,x=1
Find the zeros of x^2-3(2x-5)=10
The zeros of x^2-3(2x-5)=10 are x=5,x=1