Question

1)-x^2+x+56=0 2)4x^2+8x-60=0

Answer 1

Ответ:1)-x^2+x +56=0; x^2+7x+8x-56=0; x(x+7)-8(x+7)=0; (x+7)*(x-8)=0

X+7=0; x-8=0

X=-7, x= 8

Объяснение:

Answer 2

To solve these quadratic equations, we can use the quadratic formula:

For a quadratic equation of the form ax^2 + bx + c = 0, the solutions are given by:

x = (-b ± √(b^2 - 4ac)) / 2a

Let's apply this formula to each of the given equations:

1. -x^2 + x + 56 = 0

a = -1, b = 1, c = 56

x = (-(1) ± √((1)^2 - 4*(-1)56)) / 2(-1)

x = (-1 ± √(1 + 224)) / (-2)

x = (-1 ± √225) / (-2)

x = (-1 ± 15) / (-2)

So the two solutions for this equation are:

x1 = (-1 + 15) / (-2) = 14 / -2 = -7

x2 = (-1 - 15) / (-2) = -16 / -2 = 8

Therefore, the solutions for the first equation are x = -7 and x = 8.

2. 4x^2 + 8x - 60 = 0

a = 4, b = 8, c = -60

x = (-(8) ± √((8)^2 - 44(-60))) / 2*4

x = (-8 ± √(64 + 960)) / 8

x = (-8 ± √1024) / 8

x = (-8 ± 32) / 8

So the two solutions for this equation are:

x1 = (-8 + 32) / 8 = 24 / 8 = 3

x2 = (-8 - 32) / 8 = -40 / 8 = -5

Therefore, the solutions for the second equation are x = 3 and x = -5.