If it's not what You are looking for type in the equation solver your own equation and let us solve it.
48p^2+49p=0
a = 48; b = 49; c = 0;
Δ = b2-4ac
Δ = 492-4·48·0
Δ = 2401
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{2401}=49$$p_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(49)-49}{2*48}=\frac{-98}{96} =-1+1/48 $$p_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(49)+49}{2*48}=\frac{0}{96} =0 $
| 4+.75p=13 | | 7=5(v-4)+4v | | 5x^2-8=0. | | 4+2(x-1)=13 | | 3(x-1)-8=4(1x-x)+5 | | 3x+2,4=X+8,6 | | -(3a+6)=18 | | 3x+25+2x+10+2x-25+x-5=180 | | 4(2x+6)=5(x+ | | 14-9y=22 | | 4-t=4(t-1)-5 | | F(x)=3x-8;{-2,-1,0,1,2} | | 8*(w+2)=40 | | 3(x+4)-4=2(+x | | 20x+11°=8x+1° | | 3+y/5=8 | | 1/4(12x+8=20 | | D(4x10^-4)=2.3 | | 5(2y−4)−2(y+7)=2 | | 11x-5=3x+12+55 | | 2(2x – 5) = 6 | | -7p+8=-8p | | 10=0.25x-3 | | 240-x=284 | | 278=-v+218 | | -v+189=148 | | 3(2x-1)^2=4x-3 | | 7(2x-3)=14x+7 | | -3t=-2t+8 | | 8m+8=56 | | 1/9x=33 | | 79-k=4 |