What are the solutions to the equation 4x^2-80x+139=-5 ?

The solutions to the equation 4x^2-80x+139=-5 are x=18,x=2

Find the zeros of 4x^2-80x+139=-5

The zeros of 4x^2-80x+139=-5 are x=18,x=2

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4 x^{2} - 80 x + 139 = - 5

x = 18, x = 2

Solution steps

4 x^{2} - 80 x + 139 = - 5

Move

5

to the left side

4 x^{2} - 80 x + 144 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 80 ) \pm ( - 80 ) ^{2} - 4 \cdot 4 \cdot 144}{2 \cdot 4}

(- 80)^{2} - 4 \cdot 4 \cdot 144 = 64

x_{1, 2} = \frac{- ( - 80 ) \pm 64}{2 \cdot 4}

Separate the solutions

x_{1} = \frac{- ( - 80 ) + 64}{2 \cdot 4}, x_{2} = \frac{- ( - 80 ) - 64}{2 \cdot 4}

x = \frac{- ( - 80 ) + 64}{2 \cdot 4} : 18

x = \frac{- ( - 80 ) - 64}{2 \cdot 4} : 2

The solutions to the quadratic equation are:

x = 18, x = 2

q^{2} + 5 q - 36 = 0

- 5 x^{2} + 8 x = 9

L^{2} - 10 L + 24 = 0

0.15 c^{2} + 4.86072 c - 2.5284 = 0

x^{2} + 0.1 x + 4 = 0

What are the solutions to the equation 4x^2-80x+139=-5 ?
The solutions to the equation 4x^2-80x+139=-5 are x=18,x=2
Find the zeros of 4x^2-80x+139=-5
The zeros of 4x^2-80x+139=-5 are x=18,x=2