45 = x ( 14 − x ) 45 = x ⋅ 14 − x ⋅ x 45 = 14 x − x 2 − x 2 + 14 x − 45 = 0 a = − 1 ; b = 14 ; c = − 45 Δ = b 2 − 4 a c ; Δ = 1 4 2 − 4 ⋅ ( − 1 ) ⋅ ( − 45 ) = 196 − 180 = 16 x 1 = 2 a − b − Δ ; x 2 = 2 a − b + Δ x 1 = 2 ⋅ ( − 1 ) − 14 − 16 = − 2 − 14 − 4 = − 2 − 18 = 9 x 2 = 2 ⋅ ( − 1 ) − 14 + 16 = − 2 − 14 + 4 = − 2 − 10 = 5
A n s w er : x = 9 or x = 5.
45 = x ( 14 − x ) 45 = 14 x − x 2 x 2 − 14 x + 45 = 0 Δ = b 2 − 4 a c = ( − 14 ) 2 − 4 ∗ 1 ∗ 45 = 196 − 180 = 16 x 1 = 2 a − b − Δ = 2 14 − 16 = 2 14 − 4 = 2 12 = 6 x 2 = 2 a − b + Δ = 2 14 + 16 = 2 14 + 4 = 2 18 = 9
The solutions to the equation 45 = x ( 14 − x ) are x = 9 and x = 5 . We solved the equation by rearranging it into a standard quadratic form and applying the quadratic formula. This allowed us to find both possible values of x .
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