50=-16t^2+48t+210
Let's solve in ℝ: -16t2 + 48t + 210 = 50
ⓘ We subtract 50 from both sides
-16t2 + 48t + 210 -50 = 0
ⓘ Start from 210 and move back 50 on the number line: 210 -50 = 160
-16t2 + 48t + 160 = 0
Since -16 ≠ 0, divide both sides by -16: t2 -3t -10 = 0
Given the quadratic form at2 + bt + c, you can calculate the discriminant: Δ = b2 -4ac. Here a = 1, b = -3 and c = -10
Δ = (-3)2 -4·1·(-10)
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: (-3)2 = (-3)·(-3) = 9
Δ = 9 -4·(-10)
ⓘ Multiplications are done before additions and subtractions. Two minus signs cancel out, the product of two negative numbers is positive: -4·(-10) = 40
Δ = 9 + 40
ⓘ Start from 9 and move forward 40 on the number line: 9 + 40 = 49
Δ = 49
Since Δ > 0, the equation has two roots:
The quadratic formula gives t1 = -b -Δ2a = -(-3) -492·1
t1 = 3 -492
ⓘ The square root of a number is the positive number whose square is that number: 49 = 7; 72 = 49
t1 = 3 -72
ⓘ The numerator and the denominator are computed before the division: 3 -7 = -4
t1 = -42
ⓘ We simplify the fraction by dividing the numerator and the denominator by 2, their greatest common divisor: -42 = -2
t1 = -2
The quadratic formula gives t2 = -b + Δ2a = -(-3) + 492·1
t2 = 3 + 492
ⓘ The square root of a number is the positive number whose square is that number: 49 = 7; 72 = 49
t2 = 3 + 72
ⓘ The numerator and the denominator are computed before the division: 3 + 7 = 10
t2 = 102
ⓘ We simplify the fraction by dividing the numerator and the denominator by 2, their greatest common divisor: 102 = 5
t2 = 5
▶Solutions found: t = { -2, 5 }