If it's not what You are looking for type in the equation solver your own equation and let us solve it.
5u^2+25u-30=0
a = 5; b = 25; c = -30;
Δ = b2-4ac
Δ = 252-4·5·(-30)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{1225}=35$$u_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(25)-35}{2*5}=\frac{-60}{10} =-6 $$u_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(25)+35}{2*5}=\frac{10}{10} =1 $
| 10x+22=(23x-14)-(9x-4) | | 7v=9v=56 | | 4m+3m+6=41 | | 4(2g+8)=32 | | 13x+22=43x-458 | | 35x+5=5x-15 | | 2.4(x-2)=9.6 | | y+41/10=7 | | t/5/10=-3/10 | | (d+4)(d-2)=16 | | -3=k−85-3 | | 25x²-9=0 | | 8x+6=170-5x+8 | | 20-5+6x=45+2x+6x | | 228=4.75(8)+x | | 3/4(x+4)=8.25 | | 5(k-3)-6=5k-(4k-2) | | 5(4x-7)+8=3x | | 2m^2(2m^2)=0 | | 2z(5-(-4))=18 | | 10(k+5)=2(9k-4) | | 24+34x=183+31x | | 7x^2+360x-32400=0 | | x+(x+3)+(x+3×2)=264 | | -4b(1-9b)=0 | | 36=7m+5 | | -8-5x-2=20 | | X+(x/3)+x/3×2=264 | | 2x-26=x=16 | | -2(x-1)2-1=-9 | | 0.7x-4.2=1.4 | | 0=2(x)^2-16(x)+29 |