x²+3x<15 x²+3x-15<0 ......................................................................................................... now,you will solve it as if it was x²+3x-15=0 and the solutions that you will accept,must be <0.so,you have: ......................................................................................................... x²+3x-15=0 d=3²-4(1)(-15) d=9+60 d=69 so, x1=[-3+sq.rt.of69]/2 the sq.rt.of 69is not perfect.but it's bigger than -3,so the first result of x is possitive. now,you will find the second result of x,puting instead of + a - : x2=[-3-sq.rt.of69]/2.as you can see,the x2is a negative number so,this is the solution that we will accept. so,if we want x to be <0,the solutions of x are: x<[-3-sq.rt69]2 i hope that what i did is right,and that you understood... sq.rt. means square root

x^2 + 3x < 15 Subtract 15 from both sides x^2 + 3x -15 < 0 Now you have a quadratic eqation! Solve this equation, you'll get 2 roots 2.65, -5.65 since x^2 + 3x -15 < 0 so, your answer is -5.65 < x < 2.65

hello…here goes.

x^2 +3x -15<0 then complete the square x^2 +3x +(3/2)^2 -(3/2)^2 -15<0(for the third and 4th terms take the coefficient of x and then divide by 2) next,combine first 3 terms together. (x+(3/2))^2 - 17.25<0 (x+(3/2))^2<17.25 then when u square root u hv to use modulus so....modulus(x+(3/2))< square root of 17.25. next u hv to remove modulus...... u'll end up with.... -(square root of 17.25)< x+(3/2)