(a) (i) Factorise $x^2 + 7x - 30$ - Junior Cycle Mathematics - Question 9 - 2015

Using the factored form $(x + 10)(x - 3) = 0$, we can set each factor to zero:

1. $x + 10 = 0 ightarrow x = -10$
2. $x - 3 = 0 ightarrow x = 3$

Thus, the solutions are $x = 10$ or $x = -3$.

To solve the equation $2x^2 - 7x - 10 = 0$, we can either factor or use the quadratic formula. Here, we will use the quadratic formula:

x = rac{-b m{ ext{±}} ext{sqrt}(b^2 - 4ac)}{2a}

Substituting $a = 2$, $b = -7$, and $c = -10$,

x = rac{7 m{ ext{±}} ext{sqrt}((-7)^2 - 4(2)(-10))}{2(2)} = rac{7 m{ ext{±}} ext{sqrt}(49 + 80)}{4} = rac{7 m{ ext{±}} ext{sqrt}(129)}{4}

Calculating the two potential answers:

1. x ext{ (positive)} = rac{7 + ext{sqrt}(129)}{4} ≈ 4.59
2. x ext{ (negative)} = rac{7 - ext{sqrt}(129)}{4} ≈ -1.09

Thus, the solutions to two decimal places are $4.59$ and $-1.09$.