y-3=-(x+5)(1-x)
what is value of x for equation
y - 3 = -(x + 5)(1 - x)
y - 3 = -(x + 5 - x^2 - 5x)
y - 3 = -x - 5 + x^2 + 5x
y - 3 = x^2 + 5x - x - 5
y - 3 = x^2 + 4x - 5
y = x^2 + 4x - 2
In equations such as this, there is no single unique value for x.
You can use the quadratic equation to find any possible
intersections with the x axis, where y = 0.
x^2 + 4x - 2 = 0
-b ± √(b² - 4*a*c)
x = --------------------
2a
-4 ± √(4² - 4*1*(-2))
x = -------------------------
2*1
-4 ± √(16 - (-8))
x = ---------------------
2
-4 ± √(16 + 8)
x = -----------------
2
-4 ± √(24)
x = --------------
2
-4 ± 4.8989
x = ----------------
2
-4 + 4.8989 -4 - 4.8989
x = ---------------- and x = --------------
2 2
0.8989 -8.8989
x = --------- and x = ----------
2 2
x = 0.44945 and x = -4.44945