If it's not what You are looking for type in the equation solver your own equation and let us solve it.
20c^2+15c=0
a = 20; b = 15; c = 0;
Δ = b2-4ac
Δ = 152-4·20·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{225}=15$$c_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15}{2*20}=\frac{-30}{40} =-3/4 $$c_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15}{2*20}=\frac{0}{40} =0 $
| -5+n|4=-1 | | 2(y-3)-4(y-2)=5y+6+6(y-7) | | 0,5x-3=0 | | 6x-6=-27 | | P/5-3/8=3p/4+7/4 | | Y=0.2x^2+0x+3 | | 90=x+(3x+12) | | 48−m=20 | | 7x-70=3x-10 | | 4x^2=0.5 | | x(2x−5)=0 | | 10x+(100-x)11÷100=0 | | 5(x-3)-4(2x-9)=0 | | y-5y=2 | | 2^x+2^99x=4 | | 2^x+2^(99x)=2^2 | | 2^x+2^99x=2^2 | | X²+y²=49 | | 5m-6=2m+4 | | 9.5-x=2 | | (x*0.85)+x=60 | | 5m+9/2=19/2 | | F(4x)=8(4x)-3 | | F(4x)=8x-3 | | 6y3(3+4y2)=0 | | 62=2x+52 | | 62=x2+52 | | 15x+5x=0 | | 2n-2=3001 | | 2n-2=40002 | | 2n-2=130 | | (t+2)²+(t-3)=0 |